As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.
Definition in mathematics
A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.
Types of logarithms
For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three individual species logarithmic expressions:
- Natural logarithm ln a, where the base is the Euler number (e = 2.7).
- Decimal a, where the base is 10.
- Logarithm of any number b to base a>1.
Each of them is decided in a standard way, which includes simplification, reduction and subsequent reduction to one logarithm using logarithmic theorems. To obtain the correct values of logarithms, you should remember their properties and the sequence of actions when solving them.
Rules and some restrictions
In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract an even root from negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:
- the base "a" must always be Above zero, and at the same time not be equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
- if a > 0, then a b >0, it turns out that “c” must also be greater than zero.
How to solve logarithms?
For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.
Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.
To accurately determine the value of an unknown degree, you need to learn how to work with a table of degrees. It looks like this:
As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However, for larger values you will need a power table. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!
Equations and inequalities
It turns out that when certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
The following expression is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value “x” is under the logarithmic sign. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.
The most important difference between logarithmic equations and inequalities is that equations with logarithms (example - logarithm 2 x = √9) imply one or more specific numerical values in the answer, whereas when solving inequalities, they are defined as a region acceptable values, and the breakpoints of this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.
Basic theorems about logarithms
When solving primitive tasks of finding the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.
- The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
- The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case prerequisite is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
- The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
- The theorem in the form of a formula takes on next view: log a q b n = n/q log a b.
This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.
Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;
but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.
Examples of problems and inequalities
The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, however, it can be applied to every mathematical inequality or logarithmic equation certain rules. First of all, you should find out whether the expression can be simplified or reduced to a general form. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them quickly.
When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.
Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.
How to Use Logarithm Formulas: With Examples and Solutions
So, let's look at examples of using the basic theorems about logarithms.
- The property of the logarithm of a product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
- log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values out of the sign of the logarithm.
Assignments from the Unified State Exam
Logarithms are often found in entrance exams, especially many logarithmic problems in the Unified State Exam ( State exam for all school leavers). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge topics "Natural logarithms".
Examples and solutions to problems are taken from official Unified State Exam options. Let's see how such tasks are solved.
Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.
- It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
- All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.
So, we have powers of two. If you take the number from the bottom line, you can easily find the power to which you will have to raise 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 - actually, the definition of the logarithm:
The base a logarithm of x is the power to which a must be raised to get x.
Designation: log a x = b, where a is the base, x is the argument, b is what the logarithm is actually equal to.
For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). With the same success log 2 64 = 6, since 2 6 = 64.
The operation of finding the logarithm of a number to a given base is called logarithmization. So, let's add a new line 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 calculated so easily. For example, try finding 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 ad infinitum, and they are never repeated. If the logarithm turns out to be irrational, it is better to leave it that way: log 2 5, log 3 8, log 5 100.
It is important to understand that a logarithm is an expression with two variables (the base and the argument). At first, many people confuse where the basis is and where the argument is. To avoid annoying misunderstandings, just look at the picture:
Before us is nothing more than the definition of a logarithm. Remember: logarithm is a power, into which the base must be built in order to obtain an argument. It is the base that is raised to a power - it is highlighted in red in the picture. It turns out that the base is always at the bottom! I tell my students this wonderful rule at the very first lesson - and no confusion arises.
We've figured out the definition - all that remains is 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 the base must always be greater than zero. This follows from the definition of the degree rational indicator, to which the definition of a logarithm comes down.
- The base must be different from one, since one to any degree still remains one. 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 range of acceptable values(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). For example, the logarithm may well be negative: log 2 0.5 = −1, because 0.5 = 2 −1.
However, now we are considering only numerical expressions where it is not required to know the VA of the logarithm. All restrictions have already been taken into account by the authors of the problems. But when logarithmic equations and inequalities come into play, DL requirements will become mandatory. After all, the basis and argument may contain very strong constructions that do not necessarily correspond to the above restrictions.
Now let's consider general scheme calculating logarithms. It consists of three steps:
- Represent the base a and the argument x as a power with the minimum possible reason, greater than one. Along the way, it’s better to get rid of decimals;
- Solve the equation for 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 visible already in the first step. The requirement that the base be greater than one is very important: this reduces the likelihood of error and greatly simplifies the calculations. Same with decimals: if you immediately convert them to regular ones, there will be many fewer errors.
Let's see how this scheme works using specific examples:
Task. Calculate the logarithm: log 5 25
- Let's imagine the base and argument as a power of five: 5 = 5 1 ; 25 = 5 2 ;
- We received the answer: 2.
Let's create 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 imagine the base and argument as a power of two: 4 = 2 2 ; 64 = 2 6 ;
- Let's create and solve the equation:
log 4 64 = b ⇒ (2 2) b = 2 6 ⇒ 2 2b = 2 6 ⇒ 2b = 6 ⇒ b = 3 ; - We received the answer: 3.
Task. Calculate the logarithm: log 16 1
- Let's imagine the base and argument as a power of two: 16 = 2 4 ; 1 = 2 0 ;
- Let's create and solve the equation:
log 16 1 = b ⇒ (2 4) b = 2 0 ⇒ 2 4b = 2 0 ⇒ 4b = 0 ⇒ b = 0 ; - We received the answer: 0.
Task. Calculate the logarithm: log 7 14
- Let's imagine the base and argument as a power of seven: 7 = 7 1 ; 14 cannot be represented as a power of seven, since 7 1< 14 < 7 2 ;
- From the previous paragraph it follows that the logarithm does not count;
- The answer is no change: log 7 14.
A small note to last example. How can you be sure that a number is not an exact power of another number? It’s very simple - just factor it into prime factors. If the expansion has at least two different factors, the number is not an exact power.
Task. Find out whether the numbers are exact powers: 8; 48; 81; 35; 14 .
8 = 2 · 2 · 2 = 2 3 - exact degree, because there is only one multiplier;
48 = 6 · 8 = 3 · 2 · 2 · 2 · 2 = 3 · 2 4 - is not an exact power, since there are two factors: 3 and 2;
81 = 9 · 9 = 3 · 3 · 3 · 3 = 3 4 - exact degree;
35 = 7 · 5 - again not an exact power;
14 = 7 · 2 - again not an exact degree;
Let us also note that we ourselves prime numbers are always exact degrees of themselves.
Decimal logarithm
Some logarithms are so common that they have a special name and symbol.
The decimal logarithm of x is the logarithm to base 10, i.e. The power to which the number 10 must be raised to obtain 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 a textbook, know that this is not a typo. This is a decimal logarithm. However, if you are unfamiliar with this notation, you can always rewrite it:
log x = log 10 x
Everything that is true for ordinary logarithms is also true for decimal logarithms.
Natural logarithm
There is another logarithm that has its own designation. In some ways, it's even more important than decimal. We are talking about the natural logarithm.
The natural logarithm of x is the logarithm to base e, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x .
Many will ask: what is the number e? This irrational number, his exact value impossible to find and record. I will give only the first figures:
e = 2.718281828459...
We won’t go into detail about 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 ; ln 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, for one: ln 1 = 0.
For natural logarithms, all the rules that are true for ordinary logarithms are valid.
log a r b r =log a b or log a b= log a r b r
The value of the logarithm will not change if the base of the logarithm and the number under the logarithm sign are raised to the same power.
Under the logarithm sign there can only be positive numbers, and the base of the logarithm is not equal to one.
Examples.
1) Compare log 3 9 and log 9 81.
log 3 9=2, since 3 2 =9;
log 9 81=2, since 9 2 =81.
So log 3 9=log 9 81.
Note that the base of the second logarithm is equal to the square of the base of the first logarithm: 9=3 2, and the number under the sign of the second logarithm is equal to the square of the number under the sign of the first logarithm: 81=9 2. It turns out that both the number and the base of the first logarithm log 3 9 were raised to the second power, and the value of the logarithm did not change from this:
Next, since extracting the root n th degree from among A is the raising of a number A to the degree ( 1/n), then from log 9 81 you can get log 3 9 by taking the square root of the number and from the base of the logarithm:
2) Check equality: log 4 25=log 0.5 0.2.
Let's look at the first logarithm. Let's extract Square root from the base 4 and from among 25 ; we get: log 4 25=log 2 5.
Let's look at the second logarithm. Logarithm base: 0.5= 1 / 2. The number under the sign of this logarithm: 0.2= 1/5. Let's raise each of these numbers to the minus first power:
0,5 -1 =(1 / 2) -1 =2;
0,2 -1 =(1 / 5) -1 =5.
So log 0.5 0.2=log 2 5. Conclusion: this equality is true.
Solve the equation:
log 4 x 4 +log 16 81=log 2 (5x+2). Let's reduce logarithms from the left to the base 2 .
log 2 x 2 +log 2 3=log 2 (5x+2). Take the square root of the number and the base of the first logarithm. Extract the fourth root of the number and the base of the second logarithm.
log 2 (3x 2)=log 2 (5x+2). Convert the sum of logarithms into the logarithm of the product.
3x 2 =5x+2. Received after potentiation.
3x 2 -5x-2=0. Let's decide quadratic equation By general formula for a complete quadratic equation:
a=3, b=-5, c=-2.
D=b 2 -4ac=(-5) 2 -4∙3∙(-2)=25+24=49=7 2 >0; 2 real roots.
Examination.
x=2.
log 4 2 4 +log 16 81=log 2 (5∙2+2);
log 2 2 2 +log 2 3=log 2 12;
log 2 (4∙3)=log 2 12;
log 2 12=log 2 12;
log a n b=(1/
n)∙
log a b
Logarithm of a number b based on a n equal to the product of the fraction 1/ n to the logarithm of a number b based on a.
Find:1) 21log 8 3+40log 25 2; 2) 30log 32 3∙log 125 2 , if it is known that log 2 3=b,log 5 2=c.
Solution.
Solve equations:
1) log 2 x+log 4 x+log 16 x=5.25.
Solution.
Let's reduce these logarithms to base 2. Apply the formula: log a n b=(1/ n)∙ log a b
log 2 x+(½) log 2 x+(¼) log 2 x=5.25;
log 2 x+0.5log 2 x+0.25log 2 x=5.25. Here are similar terms:
(1+0.5+0.25) log 2 x=5.25;
1.75 log 2 x=5.25 |:1.75
log 2 x=3. By definition of logarithm:
2) 0.5log 4 (x-2)+log 16 (x-3)=0.25.
Solution. Let's convert the logarithm to base 16 to base 4.
0.5log 4 (x-2)+0.5log 4 (x-3)=0.25 |:0.5
log 4 (x-2)+log 4 (x-3)=0.5. Let's convert the sum of logarithms into the logarithm of the product.
log 4 ((x-2)(x-3))=0.5;
log 4 (x 2 -2x-3x+6)=0.5;
log 4 (x 2 -5x+6)=0.5. By definition of logarithm:
x 2 -5x+4=0. According to Vieta's theorem:
x 1 =1; x 2 =4. The first value of x will not work, since at x = 1 the logarithms of this equality do not exist, because Only positive numbers can be under the logarithm sign.
Let's check this equation at x=4.
Examination.
0.5log 4 (4-2)+log 16 (4-3)=0.25
0.5log 4 2+log 16 1=0.25
0,5∙0,5+0=0,25
log a b=log c b/log c a
Logarithm of a number b based on A equal to the logarithm of the number b on a new basis With, divided by the logarithm of the old base A on a new basis With.
Examples:
1) log 2 3=lg3/lg2;
2) log 8 7=ln7/ln8.
Calculate:
1) log 5 7, if it is known that lg7≈0,8451; lg5≈0,6990.
c b / log c a.
log 5 7=log7/log5≈0.8451:0.6990≈1.2090.
Answer: log 5 7≈1,209 0≈1,209 .
2) log 5 7 , if it is known that ln7≈1,9459; ln5≈1,6094.
Solution. Apply the formula: log a b =log c b / log c a.
log 5 7=ln7/ln5≈1.9459:1.6094≈1.2091.
Answer: log 5 7≈1,209 1≈1,209 .
Find x:
1) log 3 x=log 3 4+log 5 6/log 5 3+log 7 8/log 7 3.
We use the formula: log c b / log c a = log a b . We get:
log 3 x=log 3 4+log 3 6+log 3 8;
log 3 x=log 3 (4∙6∙8);
log 3 x=log 3 192;
x=192 .
2) log 7 x=lg143-log 6 11/log 6 10-log 5 13/log 5 10.
We use the formula: log c b / log c a = log a b . We get:
log 7 x=lg143-lg11-lg13;
log 7 x=lg143- (lg11+lg13);
log 7 x=lg143-lg (11∙13);
log 7 x=lg143-lg143;
x=1.
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The basic properties of the logarithm, logarithm graph, domain of definition, set of values, basic formulas, increasing and decreasing are given. Finding the derivative of a logarithm is considered. And also the integral, expansion in power series and representation using complex numbers.
Definition of logarithm
Logarithm with base a is a function of y (x) = log a x, inverse to the exponential function with base a: x (y) = a y.
Decimal logarithm is the logarithm to the base of a number 10 : log x ≡ log 10 x.
Natural logarithm is the logarithm to the base of e: ln x ≡ log e x.
2,718281828459045...
;
.
The graph of the logarithm is obtained from the graph of the exponential function by mirroring it with respect to the straight line y = x. On the left are graphs of the function y(x) = log a x for four values logarithm bases 2
: a = 8
: a = 1/2
, a = 1/8
and a = 1
. 0
< a < 1
the logarithm decreases monotonically.
Properties of the logarithm
Domain, set of values, increasing, decreasing
The logarithm is a monotonic function, so it has no extrema. The main properties of the logarithm are presented in the table.
| Domain | 0 < x < + ∞ | 0 < x < + ∞ |
| Range of values | - ∞ < y < + ∞ | - ∞ < y < + ∞ |
| Monotone | monotonically increases | monotonically decreases |
| Zeros, y = 0 | x = 1 | x = 1 |
| Intercept points with the ordinate axis, x = 0 | No | No |
| + ∞ | - ∞ | |
| - ∞ | + ∞ |
Private values
The logarithm to base 10 is called decimal logarithm and is denoted as follows:
Logarithm to base e called natural logarithm:
Basic formulas for logarithms
Properties of the logarithm arising from the definition of the inverse function:
The main property of logarithms and its consequences
Base replacement formula
Logarithm- This mathematical operation taking the logarithm. When taking logarithms, products of factors are converted into sums of terms.
Potentiation is the inverse mathematical operation of logarithm. During potentiation, a given base is raised to the degree of expression over which potentiation is performed. In this case, the sums of terms are transformed into products of factors.
Proof of basic formulas for logarithms
Formulas related to logarithms follow from formulas for exponential functions and from the definition of an inverse function.
Consider the property of the exponential function
.
Then
.
Let's apply the property of the exponential function
:
.
Let us prove the base replacement formula.
;
.
Assuming c = b, we have:
Inverse function
The inverse of a logarithm to base a is an exponential function with exponent a.
If , then
If , then
Derivative of logarithm
Derivative of the logarithm of modulus x:
.
Derivative of nth order:
.
Deriving formulas > > >
To find the derivative of a logarithm, it must be reduced to the base e.
;
.
Integral
The integral of the logarithm is calculated by integrating by parts: .
So,
Expressions using complex numbers
Consider the complex number function z:
.
Let's express complex number z via module r and argument φ
:
.
Then, using the properties of the logarithm, we have:
.
Or
However, the argument φ
not uniquely defined. If you put
, where n is an integer,
then it will be the same number for different n.
Therefore, the logarithm, as a function of a complex variable, is not a single-valued function.
Power series expansion
When the expansion takes place:
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.