The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is not defined. Also, the base of the logarithm must be a positive number that is not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the base -2 logarithm of 4 is 2.
Basic logarithmic identity
a log a b = b (a > 0, a ≠ 1) (2)It is important that the domains of definition of the right and left parts of this formula are different. The left side is defined only for b>0, a>0 and a ≠ 1. The right side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic "identity" in solving equations and inequalities can lead to a change in the DPV.
Two obvious consequences of the definition of the logarithm
log a a = 1 (a > 0, a ≠ 1) (3)log a 1 = 0 (a > 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the zero power, we get one.
The logarithm of the product and the logarithm of the quotient
log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against the thoughtless use of these formulas when solving logarithmic equations and inequalities. When they are used "from left to right", the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x) , we are forced to restrict ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of admissible values, and this is categorically unacceptable, since it can lead to the loss of solutions. A similar problem exists for formula (6).
The degree can be taken out of the sign of the logarithm
log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! Taking the power out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of admissible values. All these remarks apply not only to the power of 2, but also to any even power.
Formula for moving to a new base
log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)That rare case when the ODZ does not change during the conversion. If you have chosen the base c wisely (positive and not equal to 1), the formula for moving to a new base is perfectly safe.
If we choose the number b as a new base c, we obtain an important particular case of formula (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1 Calculate: lg2 + lg50.
Decision. lg2 + lg50 = lg100 = 2. We used the formula for the sum of logarithms (5) and the definition of the decimal logarithm.
Example 2 Calculate: lg125/lg5.
Decision. lg125/lg5 = log 5 125 = 3. We used the new base transition formula (8).
Table of formulas related to logarithms
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
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 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.
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 decimal system calculus it is convenient to use the table of logarithms in 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 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).
So, we have powers of two. If you take the number from the bottom line, then you can easily find the power to which you have to raise a two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.
And now - in fact, the definition of the logarithm:
The logarithm to the base a of the argument x is the power to which the number a must be raised to get the number x .
Notation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is equal to.
For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). Might as well log 2 64 = 6 because 2 6 = 64 .
The operation of finding the logarithm of a number to a given base is called the logarithm. So let's add a new row to our table:
| 2 1 | 2 2 | 2 3 | 2 4 | 2 5 | 2 6 |
| 2 | 4 | 8 | 16 | 32 | 64 |
| log 2 2 = 1 | log 2 4 = 2 | log 2 8 = 3 | log 2 16 = 4 | log 2 32 = 5 | log 2 64 = 6 |
Unfortunately, not all logarithms are considered so easily. For example, try to find log 2 5 . The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.
Such numbers are called irrational: the numbers after the decimal point can be written indefinitely, and they never repeat. If the logarithm turns out to be irrational, it is better to leave it like this: log 2 5 , log 3 8 , log 5 100 .
It is important to understand that the logarithm is an expression with two variables (base and argument). At first, many people confuse where the base is and where the argument is. To avoid annoying misunderstandings, just take a look at the picture:
Before us is nothing more than the definition of the logarithm. Remember: the logarithm is the power, to which you need to raise the base to get the argument. It is the base that is raised to a power - in the picture it is highlighted in red. It turns out that the base is always at the bottom! I tell this wonderful rule to my students at the very first lesson - and there is no confusion.
We figured out the definition - it remains to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:
- The argument and base must always be greater than zero. This follows from the definition of the degree by a rational exponent, to which the definition of the logarithm is reduced.
- The base must be different from unity, since a unit to any power is still a unit. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!
Such restrictions are called valid range(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒ x > 0 , a > 0 , a ≠ 1 .
Note that there are no restrictions on the number b (the value of the logarithm) is not imposed. For example, the logarithm may well be negative: log 2 0.5 \u003d -1, because 0.5 = 2 −1 .
However, for now we are only considering numeric expressions, where it is not required to know the ODZ of the logarithm. All restrictions have already been taken into account by the compilers of the problems. But when they go logarithmic equations and inequalities, DHS requirements will become mandatory. Indeed, in the basis and argument there can be very strong constructions, which do not necessarily correspond to the above restrictions.
Now consider general scheme logarithm calculations. It consists of three steps:
- Express the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
- Solve the equation for the variable b: x = a b ;
- The resulting number b will be the answer.
That's all! If the logarithm turns out to be irrational, this will be seen already at the first step. The requirement that the base be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similar to decimals: if you immediately translate them into ordinary ones, there will be many times less errors.
Let's see how this scheme works with specific examples:
Task. Calculate the logarithm: log 5 25
- Let's represent the base and the argument as a power of five: 5 = 5 1 ; 25 = 52;
- Received an answer: 2.
Let's make and solve the equation:
log 5 25 = b ⇒ (5 1) b = 5 2 ⇒ 5 b = 5 2 ⇒ b = 2 ;
Task. Calculate the logarithm:
Task. Calculate the logarithm: log 4 64
- Let's represent the base and the argument as a power of two: 4 = 2 2 ; 64 = 26;
- Let's make and solve the equation:
log 4 64 = b ⇒ (2 2) b = 2 6 ⇒ 2 2b = 2 6 ⇒ 2b = 6 ⇒ b = 3 ; - Received an answer: 3.
Task. Calculate the logarithm: log 16 1
- Let's represent the base and the argument as a power of two: 16 = 2 4 ; 1 = 20;
- Let's make and solve the equation:
log 16 1 = b ⇒ (2 4) b = 2 0 ⇒ 2 4b = 2 0 ⇒ 4b = 0 ⇒ b = 0 ; - Received a response: 0.
Task. Calculate the logarithm: log 7 14
- Let's represent the base and the argument as a power of seven: 7 = 7 1 ; 14 is not represented as a power of seven, because 7 1< 14 < 7 2 ;
- It follows from the previous paragraph that the logarithm is not considered;
- The answer is no change: log 7 14.
A small note on the last example. How to make sure that a number is not an exact power of another number? Very simple - just decompose it into prime factors. If there are at least two distinct factors in the expansion, the number is not an exact power.
Task. Find out if the exact powers of the number are: 8; 48; 81; 35; fourteen .
8 \u003d 2 2 2 \u003d 2 3 - the exact degree, because there is only one multiplier;
48 = 6 8 = 3 2 2 2 2 = 3 2 4 is not an exact power because there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 \u003d 3 4 - exact degree;
35 = 7 5 - again not an exact degree;
14 \u003d 7 2 - again not an exact degree;
We also note that we prime numbers are always exact powers of themselves.
Decimal logarithm
Some logarithms are so common that they have a special name and designation.
The decimal logarithm of the x argument is the base 10 logarithm, i.e. the power to which you need to raise the number 10 to get the number x. Designation: lg x .
For example, log 10 = 1; log 100 = 2; lg 1000 = 3 - etc.
From now on, when a phrase like “Find lg 0.01” appears in the textbook, know that this is not a typo. This is the decimal logarithm. However, if you are not used to such a designation, you can always rewrite it:
log x = log 10 x
Everything that is true for ordinary logarithms is also true for decimals.
natural logarithm
There is another logarithm that has its own notation. In a sense, it is even more important than decimal. This is the natural logarithm.
The natural logarithm of x is the base e logarithm, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x .
Many will ask: what else is the number e? This is an irrational number, its exact value cannot be found and written down. Here are just the first numbers:
e = 2.718281828459...
We will not delve into what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x
Thus ln e = 1 ; log e 2 = 2 ; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number irrational. Except, of course, unity: ln 1 = 0.
For natural logarithms, all the rules that are true for ordinary logarithms are valid.
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 in known value degree and 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 point two thirds to base Square root out 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.
(from the Greek λόγος - "word", "relation" and ἀριθμός - "number") numbers b by reason a(log α b) is called such a number c, and b= a c, that is, log α b=c and b=ac are equivalent. The logarithm makes sense if a > 0, a ≠ 1, b > 0.
In other words logarithm numbers b by reason a formulated as an 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 α b, is equivalent to solving the equation a x =b.
For example:
log 2 8 = 3 because 8=2 3 .
We note that the indicated formulation of the logarithm makes it possible to immediately determine logarithm value when the number under the sign of the logarithm is a certain power of the base. Indeed, 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 degree of number.
The calculation of the logarithm is referred to logarithm. Logarithm is the mathematical operation of taking a logarithm. When taking a logarithm, the products of factors are transformed into sums of terms.
Potentiation is the mathematical operation inverse to logarithm. When potentiating, the given base is raised to the power of the expression on which the potentiation is performed. In this case, the sums of terms are transformed into the product of factors.
Quite often, real logarithms with bases 2 (binary), e Euler number e ≈ 2.718 (natural logarithm) and 10 (decimal) are used.
At this stage, it is worth considering samples of logarithms log 7 2 , ln √ 5, lg0.0001.
And the entries lg (-3), log -3 3.2, log -1 -4.3 do not make sense, since in the first of them a negative number is placed under the sign of the logarithm, in the second - a negative number in the base, and in the third - and a negative number under the sign of the logarithm and unit in the base.
Conditions for determining the logarithm.
It is worth considering separately the conditions a > 0, a ≠ 1, b > 0. definition of a logarithm. Let's consider why these restrictions are taken. This will help us with an equality of the form x = log α b, called the basic logarithmic identity, which directly follows from the definition of the logarithm given above.
Take the condition a≠1. Since one is equal to one to any power, then the equality x=log α b can only exist when b=1, but log 1 1 will be any real number. To eliminate this ambiguity, we take a≠1.
Let us prove the necessity of the condition a>0. At a=0 according to the formulation of the logarithm, can only exist when b=0. And then accordingly log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. To eliminate this ambiguity, the condition a≠0. And when a<0 we would have to reject the analysis of rational and irrational values of the logarithm, since the exponent with rational and irrational exponent is defined only for non-negative bases. It is for this reason that the condition a>0.
And the last condition b>0 follows from the inequality a>0, because x=log α b, and the value of the degree with a positive base a always positive.
Features of logarithms.
Logarithms characterized by distinctive features, which led to their widespread use to greatly facilitate painstaking calculations. In the transition "to the world of logarithms", multiplication is transformed into a much easier addition, division into subtraction, and raising to a power and taking a root are transformed into multiplication and division by an exponent, respectively.
The formulation of logarithms and a table of their values (for trigonometric functions) was first published in 1614 by the Scottish mathematician John Napier. Logarithmic tables, enlarged and detailed by other scientists, were widely used in scientific and engineering calculations, and remained relevant until electronic calculators and computers began to be used.
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