We continue to study logarithms. In this article we will talk about calculation of logarithms, this process is called logarithm. First, we will deal with the calculation of logarithms by definition. Next, consider how the values of logarithms are found using their properties. After that, we will dwell on the calculation of logarithms through initially setpoints other logarithms. Finally, let's learn how to use tables of logarithms. The whole theory is provided with examples with detailed solutions.
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Computing logarithms by definition
In the simplest cases, it is possible to quickly and easily perform finding the logarithm by definition. Let's take a closer look at how this process takes place.
Its essence is to represent the number b in the form a c , whence, by the definition of the logarithm, the number c is the value of the logarithm. That is, by definition, finding the logarithm corresponds to the following chain of equalities: log a b=log a a c =c .
So, the calculation of the logarithm, by definition, comes down to finding such a number c that a c \u003d b, and the number c itself is the desired value of the logarithm.
Given the information of the previous paragraphs, when the number under the sign of the logarithm is given by some degree of the base of the logarithm, then you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show examples.
Example.
Find log 2 2 −3 , and also calculate the natural logarithm of e 5.3 .
Solution.
The definition of the logarithm allows us to say right away that log 2 2 −3 = −3 . Indeed, the number under the sign of the logarithm is equal to the base 2 to the −3 power.
Similarly, we find the second logarithm: lne 5.3 =5.3.
Answer:
log 2 2 −3 = −3 and lne 5.3 =5.3 .
If the number b under the sign of the logarithm is not given as the power of the base of the logarithm, then you need to carefully consider whether it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the sign of the logarithm is equal to the base to the power of 1, or 2, or 3, ...
Example.
Compute the logarithms log 5 25 , and .
Solution.
It is easy to see that 25=5 2 , this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2 .
We proceed to the calculation of the second logarithm. A number can be represented as a power of 7: 
(see if necessary). Hence, 
.
Let's rewrite the third logarithm in the following form. Now you can see that 
, whence we conclude that 
. Therefore, by the definition of the logarithm 
.
Briefly, the solution could be written as follows:
Answer:
log 5 25=2 , 
And .
When there is a sufficiently large value under the sign of the logarithm natural number, then it does not hurt to decompose it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore, to calculate this logarithm by definition.
Example.
Find the value of the logarithm.
Solution.
Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of one and the property of the logarithm of a number equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1 . That is, when the number 1 or the number a is under the sign of the logarithm, equal to the base of the logarithm, then in these cases the logarithms are 0 and 1, respectively.
Example.
What are the logarithms and lg10 ?
Solution.
Since , it follows from the definition of the logarithm 
.
In the second example, the number 10 under the sign of the logarithm is the same as its base, so the decimal logarithm of ten equal to one, that is, lg10=lg10 1 =1 .
Answer:
AND lg10=1 .
Note that computing logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p , which is one of the properties of logarithms.
In practice, when the number under the sign of the logarithm and the base of the logarithm are easily represented as a power of some number, it is very convenient to use the formula 
, which corresponds to one of the properties of logarithms. Consider an example of finding the logarithm, illustrating the use of this formula.
Example.
Calculate the logarithm of .
Solution.
Answer:
.
The properties of logarithms not mentioned above are also used in the calculation, but we will talk about this in the following paragraphs.
Finding logarithms in terms of other known logarithms
The information in this paragraph continues the topic of using the properties of logarithms in their calculation. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's take an example for clarification. Let's say we know that log 2 3≈1.584963 , then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .
In the above example, it was enough for us to use the property of the logarithm of the product. However, much more often you have to use a wider arsenal of properties of logarithms in order to calculate the original logarithm in terms of the given ones.
Example.
Calculate the logarithm of 27 to base 60 if it is known that log 60 2=a and log 60 5=b .
Solution.
So we need to find log 60 27 . It is easy to see that 27=3 3 , and the original logarithm, due to the property of the logarithm of the degree, can be rewritten as 3·log 60 3 .
Now let's see how log 60 3 can be expressed in terms of known logarithms. The property of the logarithm of a number equal to the base allows you to write the equality log 60 60=1 . On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2 log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2 log 60 2−log 60 5=1−2 a−b.
Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3 (1−2 a−b)=3−6 a−3 b.
Answer:
log 60 27=3 (1−2 a−b)=3−6 a−3 b.
Separately, it is worth mentioning the meaning of the formula for the transition to a new base of the logarithm of the form 
. It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, according to the transition formula, they switch to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow them to be calculated with a certain degree of accuracy. In the next section, we will show how this is done.
Tables of logarithms, their use
For an approximate calculation of the values of the logarithms, one can use logarithm tables. The most commonly used are the base 2 logarithm table, the natural logarithm table, and the decimal logarithm table. When working in the decimal number system, it is convenient to use a table of logarithms to base ten. With its help, we will learn to find the values of logarithms.
The presented table allows, with an accuracy of one ten-thousandth, to find the values of the decimal logarithms of numbers from 1.000 to 9.999 (with three decimal places). The principle of finding the value of the logarithm using the table of decimal logarithms will be analyzed in specific example- so much clearer. Let's find lg1,256 .
In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (number 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (number 6) is found in the first or last line to the right of the double line (this number is circled in green). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value of the decimal logarithm up to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.
Is it possible, using the above table, to find the values of the decimal logarithms of numbers that have more than three digits after the decimal point, and also go beyond the limits from 1 to 9.999? Yes, you can. Let's show how this is done with an example.
Let's calculate lg102.76332 . First you need to write number in standard form: 102.76332=1.0276332 10 2 . After that, the mantissa should be rounded up to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm of the resulting number, that is, we take lg102.76332≈lg1.028·10 2 . Now apply the properties of the logarithm: lg1.028 10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 according to the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the whole process of calculating the logarithm looks like this: lg102.76332=lg1.0276332 10 2 ≈lg1.028 10 2 = lg1.028+lg10 2 =lg1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using the table of decimal logarithms, you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for the transition to a new base of the logarithm, we have . From the table of decimal logarithms we find lg3≈0.4771 and lg2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).
\(a^(b)=c\) \(\Leftrightarrow\) \(\log_(a)(c)=b\)
Let's explain it easier. For example, \(\log_(2)(8)\) is equal to the power \(2\) must be raised to to get \(8\). From this it is clear that \(\log_(2)(8)=3\).
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Examples: |
\(\log_(5)(25)=2\) |
because \(5^(2)=25\) |
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\(\log_(3)(81)=4\) |
because \(3^(4)=81\) |
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\(\log_(2)\)\(\frac(1)(32)\) \(=-5\) |
because \(2^(-5)=\)\(\frac(1)(32)\) |
Argument and base of the logarithm
Any logarithm has the following "anatomy":
The argument of the logarithm is usually written at its level, and the base is written in subscript closer to the sign of the logarithm. And this entry is read like this: "the logarithm of twenty-five to the base of five."
How to calculate the logarithm?
To calculate the logarithm, you need to answer the question: to what degree should the base be raised to get the argument?
For example, calculate the logarithm: a) \(\log_(4)(16)\) b) \(\log_(3)\)\(\frac(1)(3)\) c) \(\log_(\sqrt (5))(1)\) d) \(\log_(\sqrt(7))(\sqrt(7))\) e) \(\log_(3)(\sqrt(3))\)
a) To what power must \(4\) be raised to get \(16\)? Obviously the second. That's why:
\(\log_(4)(16)=2\)
\(\log_(3)\)\(\frac(1)(3)\) \(=-1\)
c) To what power must \(\sqrt(5)\) be raised to get \(1\)? And what degree makes any number a unit? Zero, of course!
\(\log_(\sqrt(5))(1)=0\)
d) To what power must \(\sqrt(7)\) be raised to get \(\sqrt(7)\)? In the first - any number in the first degree is equal to itself.
\(\log_(\sqrt(7))(\sqrt(7))=1\)
e) To what power must \(3\) be raised to get \(\sqrt(3)\)? From we know that is a fractional power, which means Square root is the degree \(\frac(1)(2)\) .
\(\log_(3)(\sqrt(3))=\)\(\frac(1)(2)\)
Example : Calculate the logarithm \(\log_(4\sqrt(2))(8)\)
Solution :
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\(\log_(4\sqrt(2))(8)=x\) |
We need to find the value of the logarithm, let's denote it as x. Now let's use the definition of the logarithm: |
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\((4\sqrt(2))^(x)=8\) |
What links \(4\sqrt(2)\) and \(8\)? Two, because both numbers can be represented by twos: |
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\(((2^(2)\cdot2^(\frac(1)(2))))^(x)=2^(3)\) |
On the left, we use the degree properties: \(a^(m)\cdot a^(n)=a^(m+n)\) and \((a^(m))^(n)=a^(m\cdot n)\) |
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\(2^(\frac(5)(2)x)=2^(3)\) |
The bases are equal, we proceed to the equality of indicators |
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\(\frac(5x)(2)\) \(=3\) |
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Multiply both sides of the equation by \(\frac(2)(5)\) |
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The resulting root is the value of the logarithm |
Answer : \(\log_(4\sqrt(2))(8)=1,2\)
Why was the logarithm invented?
To understand this, let's solve the equation: \(3^(x)=9\). Just match \(x\) to make the equality work. Of course, \(x=2\).
Now solve the equation: \(3^(x)=8\). What is x equal to? That's the point.
The most ingenious will say: "X is a little less than two." How exactly is this number to be written? To answer this question, they came up with the logarithm. Thanks to him, the answer here can be written as \(x=\log_(3)(8)\).
I want to emphasize that \(\log_(3)(8)\), as well as any logarithm is just a number. Yes, it looks unusual, but it is short. Because if we wanted to write it as a decimal, it would look like this: \(1.892789260714.....\)
Example : Solve the equation \(4^(5x-4)=10\)
Solution :
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\(4^(5x-4)=10\) |
\(4^(5x-4)\) and \(10\) cannot be reduced to the same base. So here you can not do without the logarithm. Let's use the definition of the logarithm: |
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\(\log_(4)(10)=5x-4\) |
Flip the equation so x is on the left |
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\(5x-4=\log_(4)(10)\) |
Before us. Move \(4\) to the right. And don't be afraid of the logarithm, treat it like a regular number. |
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\(5x=\log_(4)(10)+4\) |
Divide the equation by 5 |
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\(x=\)\(\frac(\log_(4)(10)+4)(5)\) |
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Here is our root. Yes, it looks unusual, but the answer is not chosen. |
Answer : \(\frac(\log_(4)(10)+4)(5)\)
Decimal and natural logarithms
As stated in the definition of the logarithm, its base can be any positive number except one \((a>0, a\neq1)\). And among all the possible bases, there are two that occur so often that a special short notation was invented for logarithms with them:
Natural logarithm: a logarithm whose base is the Euler number \(e\) (equal to approximately \(2.7182818…\)), and the logarithm is written as \(\ln(a)\).
That is, \(\ln(a)\) is the same as \(\log_(e)(a)\)
Decimal logarithm: A logarithm whose base is 10 is written \(\lg(a)\).
That is, \(\lg(a)\) is the same as \(\log_(10)(a)\), where \(a\) is some number.
Basic logarithmic identity
Logarithms have many properties. One of them is called "Basic logarithmic identity" and looks like this:
| \(a^(\log_(a)(c))=c\) |
This property follows directly from the definition. Let's see how exactly this formula appeared.
Recall the short definition of the logarithm:
if \(a^(b)=c\), then \(\log_(a)(c)=b\)
That is, \(b\) is the same as \(\log_(a)(c)\). Then we can write \(\log_(a)(c)\) instead of \(b\) in the formula \(a^(b)=c\) . It turned out \(a^(\log_(a)(c))=c\) - the main logarithmic identity.
You can find the rest of the properties of logarithms. With their help, you can simplify and calculate the values of expressions with logarithms, which are difficult to calculate directly.
Example : Find the value of the expression \(36^(\log_(6)(5))\)
Solution :
Answer : \(25\)
How to write a number as a logarithm?
As mentioned above, any logarithm is just a number. The converse is also true: any number can be written as a logarithm. For example, we know that \(\log_(2)(4)\) is equal to two. Then you can write \(\log_(2)(4)\) instead of two.
But \(\log_(3)(9)\) is also equal to \(2\), so you can also write \(2=\log_(3)(9)\) . Similarly with \(\log_(5)(25)\), and with \(\log_(9)(81)\), etc. That is, it turns out
\(2=\log_(2)(4)=\log_(3)(9)=\log_(4)(16)=\log_(5)(25)=\log_(6)(36)=\ log_(7)(49)...\)
Thus, if we need, we can write the two as a logarithm with any base anywhere (even in an equation, even in an expression, even in an inequality) - we just write the squared base as an argument.
It's the same with a triple - it can be written as \(\log_(2)(8)\), or as \(\log_(3)(27)\), or as \(\log_(4)(64) \) ... Here we write the base in the cube as an argument:
\(3=\log_(2)(8)=\log_(3)(27)=\log_(4)(64)=\log_(5)(125)=\log_(6)(216)=\ log_(7)(343)...\)
And with four:
\(4=\log_(2)(16)=\log_(3)(81)=\log_(4)(256)=\log_(5)(625)=\log_(6)(1296)=\ log_(7)(2401)...\)
And with minus one:
\(-1=\) \(\log_(2)\)\(\frac(1)(2)\) \(=\) \(\log_(3)\)\(\frac(1)( 3)\) \(=\) \(\log_(4)\)\(\frac(1)(4)\) \(=\) \(\log_(5)\)\(\frac(1 )(5)\) \(=\) \(\log_(6)\)\(\frac(1)(6)\) \(=\) \(\log_(7)\)\(\frac (1)(7)\)\(...\)
And with one third:
\(\frac(1)(3)\) \(=\log_(2)(\sqrt(2))=\log_(3)(\sqrt(3))=\log_(4)(\sqrt( 4))=\log_(5)(\sqrt(5))=\log_(6)(\sqrt(6))=\log_(7)(\sqrt(7))...\)
Any number \(a\) can be represented as a logarithm with base \(b\): \(a=\log_(b)(b^(a))\)
Example : Find the value of an expression \(\frac(\log_(2)(14))(1+\log_(2)(7))\)
Solution :
Answer : \(1\)
Definition of logarithm
The logarithm of the number b to the base a is the exponent to which you need to raise a to get b.
The number e in mathematics, it is customary to denote the limit to which the expression tends to
Number e is irrational number- a number incommensurable with one, it cannot be exactly expressed either as a whole or as a fraction rational number.
Letter e- first letter Latin word exonere- to flaunt, hence the name in mathematics exponential- exponential function.
Number e widely used in mathematics, and in all sciences, one way or another using mathematical calculations for their needs.
Logarithms. Properties of logarithms
Definition: Logarithm positive number b in base is the exponent c to which you must raise the number a to get the number b.
Basic logarithmic identity:
7) Formula for transition to a new base:
lna = log e a, e ≈ 2.718…
Tasks and tests on the topic “Logarithms. Properties of logarithms»
- Logarithms - Important topics for repeating the exam in mathematics
To successfully complete tasks on this topic, you must know the definition of the logarithm, the properties of logarithms, the basic logarithmic identity, the definitions of decimal and natural logarithms. The main types of tasks on this topic are tasks for calculating and converting logarithmic expressions. Let's consider their solution on the following examples.
Solution: Using the properties of logarithms, we get
Solution: using the properties of the degree, we get
1) (2 2) log 2 5 =(2 log 2 5) 2 =5 2 =25
Properties of logarithms, formulations and proofs.
Logarithms have a number of characteristic properties. In this article, we will analyze the main properties of logarithms. Here we give their formulations, write down the properties of logarithms in the form of formulas, show examples of their application, and also give proofs of the properties of logarithms.
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Basic properties of logarithms, formulas
For ease of remembering and using, we present basic properties of logarithms as a list of formulas. In the next section, we give their formulations, proofs, examples of use, and necessary explanations.
and the property of the logarithm of the product of n positive numbers: log a (x 1 x 2 ... x n) \u003d log a x 1 + log a x 2 + ... + log a x n, a>0, a≠1, x 1 >0, x 2 >0, …, xn >0 .
, where a>0 , a≠1 , x>0 , y>0 .
, where a>0 , a≠1 , n is a natural number greater than one, b>0 .
, a>0 , a≠1 , b>0 , b≠1 .
, a>0 , a≠1 , b>0 , p and q are real numbers, q≠0 , in particular, for b=a we have 
.Statements and proofs of properties
We pass to the formulation and proof of the recorded properties of logarithms. All properties of logarithms are proved on the basis of the definition of the logarithm and the basic logarithmic identity that follows from it, as well as the properties of the degree.
Let's start with properties of the logarithm of unity. Its formulation is as follows: the logarithm of unity zero, that is, log a 1=0 for any a>0 , a≠1 . The proof is straightforward: since a 0 =1 for any a that satisfies the above conditions a>0 and a≠1 , then the proven equality log a 1=0 immediately follows from the definition of the logarithm.
Let's give examples of application of the considered property: log 3 1=0 , lg1=0 and .
Let's move on to the next property: the logarithm of a number equal to the base is equal to one, that is, log a a=1 for a>0 , a≠1 . Indeed, since a 1 =a for any a , then by the definition of the logarithm log a a=1 .
Examples of using this property of logarithms are log 5 5=1 , log 5.6 5.6 and lne=1 .
The logarithm of the power of a number equal to the base of the logarithm is equal to the exponent. This property of the logarithm corresponds to a formula of the form log a a p =p, where a>0 , a≠1 and p is any real number. This property follows directly from the definition of the logarithm. Note that it allows you to immediately specify the value of the logarithm, if it is possible to represent the number under the sign of the logarithm as a degree of base, we will talk more about this in the article calculating logarithms.
For example, log 2 2 7 =7 , log10 -4 =-4 and 
.
Logarithm of the product of two positive numbers x and y is equal to the product of the logarithms of these numbers: log a (x y)=log a x+log a y, a>0 , a≠1 . Let us prove the property of the logarithm of the product. Due to the properties of the degree a log a x + log a y =a log a x a log a y , and since by the main logarithmic identity a log a x =x and a log a y =y , then a log a x a log a y =x y . Thus, a log a x+log a y =x y , whence the required equality follows by the definition of the logarithm.
Let's show examples of using the property of the logarithm of the product: log 5 (2 3)=log 5 2+log 5 3 and 
.
The product logarithm property can be generalized to the product of a finite number n of positive numbers x 1 , x 2 , …, x n as log a (x 1 x 2 ... x n)= log a x 1 +log a x 2 +...+log a x n. This equality can be easily proved by the method of mathematical induction.
For example, the natural logarithm of a product can be replaced by the sum of three natural logarithms of the numbers 4 , e , and .
Logarithm of the quotient of two positive numbers x and y is equal to the difference between the logarithms of these numbers. The property of the quotient logarithm corresponds to a formula of the form , where a>0 , a≠1 , x and y are some positive numbers. The validity of this formula is proved like the formula for the logarithm of the product: since 
, then by the definition of the logarithm .
Here is an example of using this property of the logarithm: 
.
Let's move on to property of the logarithm of degree. The logarithm of a degree is equal to the product of the exponent and the logarithm of the modulus of the base of this degree. We write this property of the logarithm of the degree in the form of a formula: log a b p =p log a |b|, where a>0 , a≠1 , b and p are numbers such that the degree of b p makes sense and b p >0 .
We first prove this property for positive b . The basic logarithmic identity allows us to represent the number b as a log a b , then b p =(a log a b) p , and the resulting expression, due to the power property, is equal to a p log a b . So we arrive at the equality b p =a p log a b , from which, by the definition of the logarithm, we conclude that log a b p =p log a b .
It remains to prove this property for negative b . Here we note that the expression log a b p for negative b makes sense only for even exponents p (since the value of the degree b p must be greater than zero, otherwise the logarithm will not make sense), and in this case b p =|b| p . Then b p =|b| p =(a log a |b|) p =a p log a |b| , whence log a b p =p log a |b| .
For example, 
and ln(-3) 4 =4 ln|-3|=4 ln3 .
It follows from the previous property property of the logarithm from the root: the logarithm of the root of the nth degree is equal to the product of the fraction 1/n and the logarithm of the root expression, that is, where a>0, a≠1, n is a natural number greater than one, b>0.
The proof is based on an equality (see the definition of exponent with a fractional exponent), which is valid for any positive b , and the property of the logarithm of the degree: 
.
Here is an example of using this property: 
.
Now let's prove conversion formula to the new base of the logarithm kind 
. To do this, it suffices to prove the validity of the equality log c b=log a b log c a . The basic logarithmic identity allows us to represent the number b as a log a b , then log c b=log c a log a b . It remains to use the property of the logarithm of the degree: log c a log a b = log a b log c a . Thus, the equality log c b=log a b log c a is proved, which means that the formula for the transition to a new base of the logarithm is also proved .
Let's show a couple of examples of applying this property of logarithms: and 
.
The formula for moving to a new base allows you to move on to working with logarithms that have a “convenient” base. For example, it can be used to switch to natural or decimal logarithms so that you can calculate the value of the logarithm from a table of logarithms. The formula for the transition to a new base of the logarithm also allows in some cases to find the value of a given logarithm, when the values of some logarithms with other bases are known.
Used frequently special case formulas for the transition to a new base of the logarithm for c=b of the form . This shows that log a b and log b a are mutually inverse numbers. Eg, 
.
The formula is also often used, which is convenient when finding logarithm values. To confirm our words, we will show how the value of the logarithm of the form is calculated using it. We have 
. To prove the formula, it suffices to use the transition formula to the new base of the logarithm a: 
.
It remains to prove the comparison properties of logarithms.
Let's use the opposite method. Suppose that for a 1 >1 , a 2 >1 and a 1 2 and for 0 1 log a 1 b≤log a 2 b is true. By the properties of logarithms, these inequalities can be rewritten as 
And 
respectively, and from them it follows that log b a 1 ≤log b a 2 and log b a 1 ≥log b a 2, respectively. Then, by the properties of powers with the same bases, the equalities b log b a 1 ≥b log b a 2 and b log b a 1 ≥b log b a 2 must be satisfied, that is, a 1 ≥a 2 . Thus, we have arrived at a contradiction to the condition a 1 2 . This completes the proof.
Basic properties of logarithms
- Materials for the lesson
- Download all formulas
- log a x n = n log a x ;
Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.
These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: log a x and log a y . Then they can be added and subtracted, and:
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!
These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples - and see:
Task. Find the value of the expression: log 6 4 + log 6 9.
Since the bases of logarithms are the same, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 − log 2 3.
The bases are the same, we use the difference formula:
log 2 48 - log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 − log 3 5.
Again, the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.
As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Based on this fact, many test papers. Yes, that control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.
Removing the exponent from the logarithm
Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log 7 49 6 .
Let's get rid of the degree in the argument according to the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the value of the expression:
[Figure caption]
Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 2 4 ; 49 = 72. We have:
[Figure caption]
I think to last example clarification is required. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator have the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?
Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:
Let the logarithm log a x be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
[Figure caption]
In particular, if we put c = x , we get:
[Figure caption]
It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when deciding logarithmic equations and inequalities.
However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:
Task. Find the value of the expression: log 5 16 log 2 25.
Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let's flip the second logarithm:
[Figure caption]
Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.
Task. Find the value of the expression: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:
[Figure caption]
Now let's get rid of the decimal logarithm by moving to a new base:
[Figure caption]
Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
- n = log a a n
-
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.
The second formula is actually a paraphrased definition. It's called the basic logarithmic identity.
Indeed, what will happen if the number b is raised to such a power that the number b to this power gives the number a? That's right: this is the same number a . Read this paragraph carefully again - many people "hang" on it.
Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.
[Figure caption]
Note that log 25 64 = log 5 8 - just take the square of the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:
[Figure caption]
If someone is not in the know, this was a real task from the Unified State Examination 🙂
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.
- log a a = 1 is the logarithmic unit. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
- log a 1 = 0 is logarithmic zero. The base a can be anything, but if the argument is one - the logarithm is zero! Because a 0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out - and solve the problems.
Logarithm. Properties of the logarithm (addition and subtraction).
Properties of the logarithm follow from its definition. And so the logarithm of the number b by reason A defined as the exponent to which a number must be raised a to get the number b(the logarithm exists only for positive numbers).
From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation ax=b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b by reason a equals With. It is also clear that the topic of logarithm is closely related to the topic of the power of a number.
With logarithms, as with any numbers, you can perform addition, subtraction operations and transform in every possible way. But in view of the fact that logarithms are not quite ordinary numbers, their own special rules apply here, which are called basic properties.
Addition and subtraction of logarithms.
Take two logarithms with the same base: log x And log a y. Then remove it is possible to perform addition and subtraction operations:
As we see, sum of logarithms equals the logarithm of the product, and difference logarithms- the logarithm of the quotient. And this is true if the numbers A, X And at positive and a ≠ 1.
It is important to note that the main aspect in these formulas are the same bases. If the bases differ from each other, these rules do not apply!
The rules for adding and subtracting logarithms with the same bases are read not only from left to right, but also vice versa. As a result, we have the theorems for the logarithm of the product and the logarithm of the quotient.
Logarithm of the product two positive numbers is equal to the sum of their logarithms ; paraphrasing this theorem, we get the following, if the numbers A, x And at positive and a ≠ 1, That:
Logarithm of the quotient of two positive numbers is equal to the difference between the logarithms of the dividend and the divisor. In other words, if the numbers A, X And at positive and a ≠ 1, That:
We apply the above theorems to solve examples:
If numbers x And at are negative, then product logarithm formula becomes meaningless. So, it is forbidden to write:
since the expressions log 2 (-8) and log 2 (-4) are not defined at all ( logarithmic function at= log 2 X defined only for positive values argument X).
Product theorem is applicable not only for two, but also for an unlimited number of factors. This means that for every natural k and any positive numbers x 1 , x 2 , . . . ,x n there is an identity:
From quotient logarithm theorems one more property of the logarithm can be obtained. It is well known that log a 1= 0, therefore,
So there is an equality:
Logarithms of two mutually reciprocal numbers on the same basis will differ from each other only in sign. So:
Logarithm. Properties of logarithms
Logarithm. Properties of logarithms
Consider equality. Let us know the values and and we want to find the value of .
That is, we are looking for an exponent to which you need to cock to get .
Let
the variable can take any real value, then the following restrictions are imposed on the variables: o” title=”a>o”/> , 1″ title=”a1″/>, 0″ title=”b>0″/>If we know the values \u200b\u200bof and , and we are faced with the task of finding the unknown, then for this purpose a mathematical operation is introduced, which is called logarithm.
To find the value we take logarithm of a number By foundation :
The logarithm of a number to the base is the exponent to which you need to raise to get .
That is basic logarithmic identity:
o” title=”a>o”/> , 1″ title=”a1″/>, 0″ title=”b>0″/>
is essentially mathematical notation logarithm definitions.
The mathematical operation logarithm is the inverse of exponentiation, so properties of logarithms are closely related to the properties of the degree.
We list the main properties of logarithms:
(o” title=”a>o”/> , 1″ title=”a1″/>, 0″ title=”b>0″/>, 0,
d>0″/>, 1″ title=”d1″/>
4.
5.
The following group of properties allows you to represent the exponent of the expression under the sign of the logarithm, or standing at the base of the logarithm as a coefficient before the sign of the logarithm:
6.
7.
8.
9.
The next group of formulas allows you to go from a logarithm with a given base to a logarithm with an arbitrary base, and is called transition formulas to a new base:
10.
12. (corollary from property 11)
The following three properties are not well known, but they are often used when solving logarithmic equations, or when simplifying expressions containing logarithms:
13.
14.
15.
Special cases:
— decimal logarithm
— natural logarithmWhen simplifying expressions containing logarithms, a general approach is applied:
1. Introducing decimals in the form of ordinary.
2. mixed numbers represented as improper fractions.
3. The numbers at the base of the logarithm and under the sign of the logarithm are decomposed into prime factors.
4. We try to bring all logarithms to the same base.
5. Apply the properties of logarithms.
Let's look at examples of simplifying expressions containing logarithms.
Example 1
Calculate:
Let's simplify all the exponents: our task is to bring them to logarithms, the base of which is the same number as the base of the exponent.
==(by property 7)=(by property 6) =
Substitute the indicators that we have obtained in the original expression. We get:
Answer: 5.25
Example 2 Calculate:
We bring all logarithms to base 6 (in this case, the logarithms from the denominator of the fraction will “migrate” to the numerator):
Let's decompose the numbers under the sign of the logarithm into prime factors:
Apply properties 4 and 6:
We introduce the replacement
We get:
Answer: 1
Logarithm . Basic logarithmic identity.
Properties of logarithms. Decimal logarithm. natural logarithm.
logarithm positive number N in base (b > 0, b 1) is called the exponent x to which you need to raise b to get N .
This entry is equivalent to the following: b x = N .
EXAMPLES: log 3 81 = 4 since 3 4 = 81 ;
log 1/3 27 = – 3 because (1/3) - 3 = 3 3 = 27 .
The above definition of the logarithm can be written as an identity:
Basic properties of logarithms.
2) log 1 = 0 because b 0 = 1 .
3) The logarithm of the product is equal to the sum of the logarithms of the factors:
4) The logarithm of the quotient is equal to the difference between the logarithms of the dividend and the divisor:
5) The logarithm of the degree is equal to the product of the exponent and the logarithm of its base:
The consequence of this property is the following: log root equals the logarithm of the root number divided by the power of the root:
6) If the base of the logarithm is a power, then the value the reciprocal of the exponent can be taken out of the rhyme log sign:
The last two properties can be combined into one:
7) The formula for the transition modulus (i.e. the transition from one base of the logarithm to another base):
In a particular case, when N = a we have:
Decimal logarithm called base logarithm 10. It is denoted lg, i.e. log 10 N= log N. Logarithms of numbers 10, 100, 1000, . p are 1, 2, 3, …, respectively, i.e. have so many positive
units, how many zeros are in the logarithm number after one. Logarithms of numbers 0.1, 0.01, 0.001, . p are –1, –2, –3, …, respectively, i.e. have as many negative ones as there are zeros in the logarithm number before the one (including zero integers). The logarithms of the remaining numbers have a fractional part called mantissa. The integer part of the logarithm is called characteristic. For practical applications, decimal logarithms are most convenient.
natural logarithm called base logarithm e. It is denoted by ln, i.e. log e N=ln N. Number e is irrational, its approximate value is 2.718281828. It is the limit towards which the number (1 + 1 / n) n with unlimited increase n(cm. first wonderful limit on the Number Sequence Limits page).
Strange as it may seem, natural logarithms turned out to be very convenient when carrying out various operations related to the analysis of functions. Calculating base logarithms e much faster than any other basis.
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The focus of this article is logarithm. Here we will give the definition of the logarithm, show the accepted notation, give examples of logarithms, and talk about natural and decimal logarithms. After that, consider the basic logarithmic identity.
Page navigation.
Definition of logarithm
The concept of a logarithm arises when solving a problem in a certain sense inverse, when you need to find the exponent from a known value of the degree and a known base.
But enough preamble, it's time to answer the question "what is a logarithm"? Let us give an appropriate definition.
Definition.
Logarithm of b to base a, where a>0 , a≠1 and b>0 is the exponent to which you need to raise the number a to get b as a result.
At this stage, we note that the spoken word "logarithm" should immediately raise two ensuing questions: "what number" and "on what basis." In other words, there is simply no logarithm, but there is only the logarithm of a number in some base.
We will immediately introduce logarithm notation: the logarithm of the number b to the base a is usually denoted as log a b . The logarithm of the number b to the base e and the logarithm to the base 10 have their own special designations lnb and lgb respectively, that is, they write not log e b , but lnb , and not log 10 b , but lgb .
Now you can bring: .
And the records 
do not make sense, since in the first of them under the sign of the logarithm is a negative number, in the second - a negative number in the base, and in the third - both a negative number under the sign of the logarithm and a unit in the base.
Now let's talk about rules for reading logarithms. The entry log a b is read as "logarithm of b to base a". For example, log 2 3 is the logarithm of three to base 2, and is the logarithm of two integer two base thirds of the square root of five. The logarithm to base e is called natural logarithm, and the notation lnb is read as "the natural logarithm of b". For example, ln7 is the natural logarithm of seven, and we will read it as the natural logarithm of pi. The logarithm to base 10 also has a special name - decimal logarithm, and the notation lgb is read as "decimal logarithm b". For example, lg1 is the decimal logarithm of one, and lg2.75 is the decimal logarithm of two point seventy-five hundredths.
It is worth dwelling separately on the conditions a>0, a≠1 and b>0, under which the definition of the logarithm is given. Let us explain where these restrictions come from. To do this, we will be helped by an equality of the form, called , which directly follows from the definition of the logarithm given above.
Let's start with a≠1 . Since the unit is equal to one to any power, then the equality can be valid only for b=1, but at the same time log 1 1 can be any real number. To avoid this ambiguity, a≠1 is accepted.
Let us substantiate the expediency of the condition a>0 . With a=0, by the definition of the logarithm, we would have equality , which is possible only with b=0 . But then log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. This ambiguity can be avoided by the condition a≠0 . And for a<0 нам бы пришлось отказаться от рассмотрения рациональных и иррациональных значений логарифма, так как степень с рациональным и иррациональным показателем определена лишь для неотрицательных оснований. Поэтому и принимается условие a>0 .
Finally, the condition b>0 follows from the inequality a>0 , since , and the value of the degree with a positive base a is always positive.
In conclusion of this paragraph, we say that the voiced definition of the logarithm allows you to immediately indicate the value of the logarithm when the number under the sign of the logarithm is a certain degree of base. Indeed, the definition of the logarithm allows us to assert that if b=a p , then the logarithm of the number b to the base a is equal to p . That is, the equality log a a p =p is true. For example, we know that 2 3 =8 , then log 2 8=3 . We will talk more about this in the article.
basic properties.
- logax + logay = log(x y);
- logax − logay = log(x: y).
same grounds
log6 4 + log6 9.
Now let's complicate the task a little.
Examples of solving logarithms
What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x >
Task. Find the value of the expression:
Transition to a new foundation
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
Task. Find the value of the expression:
See also:
Basic properties of the logarithm
1.
2.
3.
4.
5. 
6.
7.
8.
9.
10.
11. 
12.
13.
14. 
15.
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy.
Basic properties of logarithms
Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.
Examples for logarithms
Take the logarithm of expressions
Example 1
A). x=10ac^2 (a>0, c>0).
By properties 3,5 we calculate 
2.
3. 
4. 
Where 
.
Example 2 Find x if
Example 3. Let the value of logarithms be given
Calculate log(x) if
Basic properties of logarithms
Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.
These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:
- logax + logay = log(x y);
- logax − logay = log(x: y).
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!
These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:
Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
Task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
Task. Find the value of the expression: log3 135 − log3 5.
Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.
Removing the exponent from the logarithm
It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
Task. Find the value of the expression:
Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:
I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator.
Formulas of logarithms. Logarithms are examples of solutions.
They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?
Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we put c = x, we get:
It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:
Task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let's flip the second logarithm:
Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.
Task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.
The second formula is actually a paraphrased definition. It's called like this:
Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.
Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the value of the expression:
Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:
If someone is not in the know, this was a real task from the Unified State Examination 🙂
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out and solve the problems.
See also:
The logarithm of the number b to the base a denotes the expression. To calculate the logarithm means to find such a power x () at which the equality is true
Basic properties of the logarithm
The above properties need to be known, since, on their basis, almost all problems and examples are solved based on logarithms. The remaining exotic properties can be derived by mathematical manipulations with these formulas
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.
Common cases of logarithms
Some of the common logarithms are those in which the base is even ten, exponential or deuce.
The base ten logarithm is usually called the base ten logarithm and is simply denoted lg(x).
It can be seen from the record that the basics are not written in the record. For example
The natural logarithm is the logarithm whose basis is the exponent (denoted ln(x)).
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.
And another important base two logarithm is
The derivative of the logarithm of the function is equal to one divided by the variable
The integral or antiderivative logarithm is determined by the dependence
The above material is enough for you to solve a wide class of problems related to logarithms and logarithms. For the sake of understanding the material, I will give only a few common examples from school curriculum and universities.
Examples for logarithms
Take the logarithm of expressions
Example 1
A). x=10ac^2 (a>0, c>0).
By properties 3,5 we calculate
2.
By the difference property of logarithms, we have
3.
Using properties 3.5 we find
4. Where .
A seemingly complex expression using a series of rules is simplified to the form
Finding Logarithm Values
Example 2 Find x if
Solution. For the calculation, we apply properties 5 and 13 up to the last term
Substitute in the record and mourn
Since the bases are equal, we equate the expressions
Logarithms. First level.
Let the value of the logarithms be given
Calculate log(x) if
Solution: Take the logarithm of the variable to write the logarithm through the sum of the terms
This is just the beginning of acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the acquired knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another equally important topic - logarithmic inequalities ...
Basic properties of logarithms
Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.
These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:
- logax + logay = log(x y);
- logax − logay = log(x: y).
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!
These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:
Task. Find the value of the expression: log6 4 + log6 9.
Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
Task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
Task. Find the value of the expression: log3 135 − log3 5.
Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.
Removing the exponent from the logarithm
Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows their first two. But it's better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. you can enter the numbers before the sign of the logarithm into the logarithm itself.
How to solve logarithms
This is what is most often required.
Task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
Task. Find the value of the expression:
Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:
I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which was done. The result is the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the bases are different? What if they are not exact powers of the same number?
Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we put c = x, we get:
It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:
Task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let's flip the second logarithm:
Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.
Task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.
The second formula is actually a paraphrased definition. It's called like this:
Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.
Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the value of the expression:
Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:
If someone is not in the know, this was a real task from the Unified State Examination 🙂
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out and solve the problems.
