derived 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 the logarithm is closely related to the topic of the power of a number.
With logarithms, as with any numbers, you can perform operations of addition, subtraction 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:
log a x+ log a y= log a (x y);
log a x - log a y = log a (x:y).
log a(x 1 . x 2 . x 3 ... x k) = log x 1 + log x 2 + log x 3 + ... + log a x k.
From quotient logarithm theorems one more property of the logarithm can be obtained. It is well known that log a 1= 0, therefore,
log a 1 /b= log a 1 - log a b= -log a b.
So there is an equality:
log a 1 / b = - log a b.
Logarithms of two mutually reciprocal numbers on the same basis will differ from each other only in sign. So:
Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.
What is a logarithm?
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
What is a logarithm? How to solve logarithms? These questions confuse many graduates. Traditionally, the topic of logarithms is considered complex, incomprehensible and scary. Especially - equations with logarithms.
This is absolutely not true. Absolutely! Don't believe? Good. Now, for some 10 - 20 minutes you:
1. Understand what is a logarithm.
2. Learn to solve a whole class of exponential equations. Even if you haven't heard of them.
3. Learn to calculate simple logarithms.
Moreover, for this you will only need to know the multiplication table, and how a number is raised to a power ...
I feel you doubt ... Well, keep time! Go!
First, solve the following equation in your mind:
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
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.
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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 there is a negative number under the sign of the logarithm, 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 one is equal to one to any power, the equality can only be true for b=1, but 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.
Today we will talk about logarithm formulas and give demonstration solution examples.
By themselves, they imply solution patterns according to the basic properties of logarithms. Before applying the logarithm formulas to the solution, we recall for you, first all the properties:
Now, based on these formulas (properties), we show examples of solving logarithms.
Examples of solving logarithms based on formulas.
Logarithm a positive number b in base a (denoted log a b) is the exponent to which a must be raised to get b, with b > 0, a > 0, and 1.
According to the definition log a b = x, which is equivalent to a x = b, so log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2 because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm is an ordinary logarithm, the base of which is 10. Denoted as lg.
log 10 100 = 2 because 10 2 = 100
natural logarithm- also the usual logarithm logarithm, but with the base e (e \u003d 2.71828 ... - an irrational number). Referred to as ln.
It is desirable to remember the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
- Basic logarithmic identity
a log a b = b8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
- The logarithm of the product is equal to the sum of the logarithms
log a (bc) = log a b + log a clog 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
- The logarithm of the quotient is equal to the difference of the logarithms
log a (b/c) = log a b - log a c9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
- Properties of the degree of a logarithmable number and the base of the logarithm
The exponent of a logarithm number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
- Transition to a new foundation
log a b = log c b / log c a,if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the logarithm formulas are not as complicated as they seem. Now, having considered examples of solving logarithms, we can move on to logarithmic equations. We will consider examples of solving logarithmic equations in more detail in the article: "". Do not miss!
If you still have questions about the solution, write them in the comments to the article.
Note: decided to get an education of another class study abroad as an option.
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 the initially given values of 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 .
Decision.
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 .
Decision.
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 a sufficiently large natural number is under the sign of the logarithm, 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.
Decision.
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 ?
Decision.
Since , it follows from the definition of the logarithm 
.
In the second example, the number 10 under the sign of the logarithm coincides with its base, so the decimal logarithm of ten is 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 .
Decision.
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 .
Decision.
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 calculating their values 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). We will analyze the principle of finding the value of the logarithm using a table of decimal logarithms using a specific example - it’s 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 the marked columns (these numbers are highlighted in 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).
