Transition to a new foundation. Logarithm formulas

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Today we will talk about logarithmic formulas and we will give indicative solution examples.

They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithm formulas to solve, let us remind you of all the properties:

Now, based on these formulas (properties), we will show examples of solving logarithms.

Examples of solving logarithms based on formulas.

Logarithm a positive number b to base a (denoted by log a b) is an exponent to which a must be raised to obtain b, with b > 0, a > 0, and 1.

According to the definition, log a b = x, which is equivalent to a x = b, therefore 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- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.

log 10 100 = 2, because 10 2 = 100

Natural logarithm- also an ordinary logarithm, a logarithm, but with the base e (e = 2.71828... - an irrational number). Denoted as ln.

It is advisable to memorize 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 = b

  8 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 c

  log 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 c

  9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81

• Properties of the power of a logarithmic number and the base of the logarithm

  Exponent of the logarithmic 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 formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at examples of solving logarithmic equations in more detail in the article: "". Don't miss it!

If you still have questions about the solution, write them in the comments to the article.

Note: we decided to get a different class of education and study abroad as an option.

As society developed and production became more complex, mathematics also developed. Movement from simple to complex. From ordinary accounting using the method of addition and subtraction, with their repeated repetition, we came to the concept of multiplication and division. Reducing the repeated operation of multiplication became the concept of exponentiation. The first tables of the dependence of numbers on the base and the number of exponentiation were compiled back in the 8th century by the Indian mathematician Varasena. From them you can count the time of occurrence of logarithms.

Historical sketch

The revival of Europe in the 16th century also stimulated the development of mechanics. T required a large amount of computation related to multiplication and division of multi-digit numbers. The ancient tables were of great service. They made it possible to replace complex operations with simpler ones - addition and subtraction. A big step forward was the work of the mathematician Michael Stiefel, published in 1544, in which he realized the idea of ​​​​many mathematicians. This made it possible to use tables not only for powers in the form of prime numbers, but also for arbitrary rational ones.

In 1614, the Scotsman John Napier, developing these ideas, first introduced the new term “logarithm of a number.” New complex tables were compiled for calculating the logarithms of sines and cosines, as well as tangents. This greatly reduced the work of astronomers.

New tables began to appear, which were successfully used by scientists for three centuries. A lot of time passed before the new operation in algebra acquired its finished form. The definition of the logarithm was given and its properties were studied.

Only in the 20th century, with the advent of the calculator and computer, did humanity abandon the ancient tables that had worked successfully throughout the 13th centuries.

Today we call the logarithm of b to base a the number x that is the power of a to make b. This is written as a formula: x = log a(b).

For example, log 3(9) will be equal to 2. This is obvious if you follow the definition. If we raise 3 to the power of 2, we get 9.

Thus, the formulated definition sets only one restriction: the numbers a and b must be real.

Types of logarithms

The classic definition is called the real logarithm and is actually the solution to the equation a x = b. Option a = 1 is borderline and is not of interest. Attention: 1 to any power is equal to 1.

Real value of logarithm defined only when the base and the argument are greater than 0, and the base must not be equal to 1.

Special place in the field of mathematics play logarithms, which will be named depending on the size of their base:

Rules and restrictions

The fundamental property of logarithms is the rule: the logarithm of a product is equal to the logarithmic sum. log abp = log a(b) + log a(p).

As a variant of this statement it will be: log c(b/p) = log c(b) - log c(p), the quotient function is equal to the difference of the functions.

From the previous two rules it is easy to see that: log a(b p) = p * log a(b).

Other properties include:

Comment. There is no need to make a common mistake - the logarithm of a sum is not equal to the sum of logarithms.

For many centuries, the operation of finding a logarithm was a rather time-consuming task. Mathematicians used the well-known formula of the logarithmic theory of polynomial expansion:

ln (1 + x) = x — (x^2)/2 + (x^3)/3 — (x^4)/4 + … + ((-1)^(n + 1))*(( x^n)/n), where n is a natural number greater than 1, which determines the accuracy of the calculation.

Logarithms with other bases were calculated using the theorem about the transition from one base to another and the property of the logarithm of the product.

Since this method is very labor-intensive and when solving practical problems difficult to implement, we used pre-compiled tables of logarithms, which significantly speeded up all the work.

In some cases, specially compiled graphs of logarithms were used, which gave less accuracy, but significantly speeded up the search for the desired value. The curve of the function y = log a(x), constructed over several points, allows you to use a regular ruler to find the value of the function at any other point. For a long time, engineers used so-called graph paper for these purposes.

In the 17th century, the first auxiliary analog computing conditions appeared, which by the 19th century acquired a complete form. The most successful device was called the slide rule. Despite the simplicity of the device, its appearance significantly accelerated the process of all engineering calculations, and this is difficult to overestimate. Currently, few people are familiar with this device.

The advent of calculators and computers made the use of any other devices pointless.

Equations and inequalities

To solve various equations and inequalities using logarithms, the following formulas are used:

• Transition from one base to another: log a(b) = log c(b) / log c(a);
• As a consequence of the previous option: log a(b) = 1 / log b(a).

To solve inequalities it is useful to know:

• The value of the logarithm will be positive only if the base and argument are both greater or less than one; if at least one condition is violated, the logarithm value will be negative.
• If the logarithm function is applied to the right and left sides of an inequality, and the base of the logarithm is greater than one, then the sign of the inequality is preserved; otherwise it changes.

Examples of problems

Let's consider several options for using logarithms and their properties. Examples with solving equations:

Consider the option of placing the logarithm in a power:

• Problem 3. Calculate 25^log 5(3). Solution: in the conditions of the problem, the entry is similar to the following (5^2)^log5(3) or 5^(2 * log 5(3)). Let's write it differently: 5^log 5(3*2), or the square of a number as a function argument can be written as the square of the function itself (5^log 5(3))^2. Using the properties of logarithms, this expression is equal to 3^2. Answer: as a result of the calculation we get 9.

Practical Application

Being a purely mathematical tool, it seems far from real life that the logarithm suddenly acquired great importance for describing objects in the real world. It is difficult to find a science where it is not used. This fully applies not only to natural, but also to humanitarian fields of knowledge.

Logarithmic dependencies

Here are some examples of numerical dependencies:

Mechanics and physics

Historically, mechanics and physics have always developed using mathematical research methods and at the same time served as an incentive for the development of mathematics, including logarithms. The theory of most laws of physics is written in the language of mathematics. Let us give only two examples of describing physical laws using the logarithm.

The problem of calculating such a complex quantity as rocket speed can be solved by using the Tsiolkovsky formula, which laid the foundation for the theory of space exploration:

V = I * ln (M1/M2), where

• V is the final speed of the aircraft.
• I – specific impulse of the engine.
• M 1 – initial mass of the rocket.
• M 2 – final mass.

Another important example- this is used in the formula of another great scientist Max Planck, which serves to evaluate the equilibrium state in thermodynamics.

S = k * ln (Ω), where

• S – thermodynamic property.
• k – Boltzmann constant.
• Ω is the statistical weight of different states.

Chemistry

Less obvious is the use of formulas in chemistry containing the ratio of logarithms. Let's give just two examples:

• Nernst equation, the condition of the redox potential of the medium in relation to the activity of substances and the equilibrium constant.
• The calculation of such constants as the autolysis index and the acidity of the solution also cannot be done without our function.

Psychology and biology

And it’s not at all clear what psychology has to do with it. It turns out that the strength of sensation is well described by this function as the inverse ratio of the stimulus intensity value to the lower intensity value.

After the above examples, it is no longer surprising that the topic of logarithms is widely used in biology. Entire volumes could be written about biological forms corresponding to logarithmic spirals.

Other areas

It seems that the existence of the world is impossible without connection with this function, and it rules all laws. Especially when the laws of nature are associated with geometric progression. It’s worth turning to the MatProfi website, and there are many such examples in the following areas of activity:

The list can be endless. Having mastered the basic principles of this function, you can plunge into the world of infinite wisdom.

Let's look at examples of logarithmic equations.

Example 1: Solve the equation

To solve this, we use the potentiation method. Inequalities >0 and >0 will determine the range of acceptable values ​​of the equation. The inequality >0 is valid for any values ​​of x, since a 5x>0 only for positive values ​​of x. This means that the ODZ equation is a set of numbers from zero to plus infinity. The equation is equivalent to a quadratic equation. The roots of this equation are the numbers 2 and 3, since the product of these numbers is equal to 6, and the sum of these numbers is equal to 5 - the opposite value of the coefficient b? Both of these numbers lie in the interval, which means they are the roots of this equation. Note that we easily solved this equation.

Example 2: Solve the equation

(the logarithm of the expression ten x minus nine to base three is equal to the logarithm of x to base one third)

This equation differs from the previous one in that the logarithms have different bases. And the considered method for solving the equation can no longer be used here, although you can find the range of acceptable values ​​and try to solve the equation using a functional graphical method. Inequalities >0 and x>0 determine the range of permissible values ​​of the equation, which means. Let's look at a graphical illustration of this equation. To do this, let's construct a point-by-point graph of the function and. We can only say that this equation has a single root, it is positive and lies in the interval from 1 to 2. It is not possible to give the exact value of the root.

Of course, this equation is not the only one containing logarithms with different bases. Such equations can only be solved by moving to a new logarithm base. Difficulties associated with logarithms of different bases can also be encountered in other types of tasks. For example, when comparing numbers and.

An assistant in solving such problems is the theorem

Theorem: If a,b,c are positive numbers, and a and c are different from 1, then the equality holds

This formula is called the formula for moving to a new base)

Thus, from and more. Since, according to the formula for moving to a new base, equal and equal

Let us prove the theorem about the transition to a new base of the logarithm.

To prove this, we introduce the notation = m, =n, =k(the logarithm of the number BE to the base a is equal to em, the logarithm of the number BE to the base CE is equal to en, the logarithm of the number a to the base CE is equal to ka). Then, by the definition of the logarithm: the number b is a to the power of m, the number b is c to the power of n, the number a is c to the power k. So, let’s substitute its value in when raising a degree to a power, the exponents of the powers are multiplied, we get that =, but therefore =, if the bases of the degree are equal, then the exponents of the given degree are equal =. So = let's return to the reverse substitution: (the logarithm of the number BE to the base a is equal to the ratio of the logarithm of the number BE to the base CE to the logarithm of the number a to the base CE)

Let us consider two corollaries of this theorem.

First consequence. Let in this theorem we want to go to the base b. Then

(logarithm of the number BE to the base BE divided by the logarithm of the number a to the base BE)

is equal to one, then it is equal to

This means that if a and b are positive numbers and different from 1, then the equality is true

Corollary 2. If a and b are positive numbers, and A number not equal to one, then for any number m, not equal to zero, the equality is true

logarithm b based on A equal to the logarithm b to a degree m based on a to a degree m.

Let us prove this equality from right to left. Let's move from the expression (logarithm of the number be to the power em to the base a to the power em) to the logarithm with the base A. By the property of the logarithm, the exponent of a sublogarithmic expression can be moved forward - in front of the logarithm. =1. We'll get it. (a fraction in the numerator em multiplied by the logarithm of the number be to the base and in the denominator em) The number m is not equal to zero by condition, which means that the resulting fraction can be reduced by m. We'll get it. Q.E.D.

This means that to move to a new base of the logarithm, three formulas are used

Example 2: Solve the equation

(the logarithm of the expression ten x minus nine to base three is equal to the logarithm of x to base one third)

We found the range of acceptable values ​​for this equation earlier. Let's bring 3 to a new base. To do this, write it in this logarithm as a fraction. The numerator will be the logarithm of x to base three, the denominator will be the logarithm of one third to base three. is equal to minus one, then the right side of the equation will be equal to minus

Let's move it to the left side of the equation and write it as: By property, the sum of logarithms is equal to the logarithm of the product, which means (the logarithm of the expression ten x minus nine to base three plus the logarithm of x to base three) can be written as. (logarithm of the product ten x minus nine and x to base three) Let’s perform the multiplication and get parts of the equation

and on the right side we will write zero as, since three to the zero power is one.

Using the potentiation method we obtain the quadratic equation =0. By the property of the coefficients a+b+c=0, the roots of the equation are equal to 1 and 0.1.

But there is only one root in the domain of definition. This is number one.

Example 3. Calculate. (three to the power of four times the logarithm of two to the base three plus the logarithm of the root of two to the base five times the logarithm of twenty-five to the base four)

First, let's look at the power of three. If powers are multiplied, then the action of raising a power to a power is performed, so the power of three can be written as three to the power of four. Logarithms in a product with different bases; it is more convenient to reduce the logarithm with base four to the base associated with five. Therefore, let us replace it with an expression identical to it. According to the formula for moving to a new base.

According to the basic logarithmic identity (and to the power, the logarithm of the number of bes to base a is equal to the number of bes)

instead we get In the expression, we select the square of the base and the sublogarithmic expression. We'll get it. According to the formula for transition to a new base, it is written to the right of the solution, we get instead of only. We write the square root of two as two to the power of one-half and, by the property of the logarithm, put the exponent in front of the logarithm. Let's get the expression. Thus, the calculated expression will take the form...

Moreover, this is 16, and the product is equal to one, which means the value of the expression is 16.5.

Example 4. Calculate if log2= a,log3= b

To calculate, we will use the properties of the logarithm and the formulas for transition to a new base.

Let's imagine 18 as the product of six and three. The logarithm of the product is equal to the sum of the logarithms-factors, that is, where is equal to 1. Since we know decimal logarithms, we move from the logarithm with base 6 to the decimal logarithm, we obtain a fraction in the numerator (decimal logarithm of three) and in the denominator (decimal logarithm of six ). In this case, you can already replace it with b. Let's factor six into factors of two and three. We write the resulting product as a sum of logarithms lg2 and lg 3. Replace them with a and b, respectively. The expression will take the form: . If this expression is converted into a fraction by reduction to a common denominator, then the answer will be

To successfully complete tasks related to the transition to a new logarithm base, you need to know the formulas for transition to a new logarithm base

1. , where a,b,c are positive numbers, a, c
2. , where a, b are positive numbers, a, b
3. , where a,b are positive numbers a, m

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 undefined. In addition, 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 logarithm to the base -2 of 4 is equal to 2.

Basic logarithmic identity

It is important that the scope of definition of the right and left sides of this formula is 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” when solving equations and inequalities can lead to a change in the OD.

Two obvious consequences of the definition of logarithm

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.

Logarithm of the product and logarithm of the quotient

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 thoughtlessly using these formulas when solving logarithmic equations and inequalities. When using them “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 limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A similar problem exists for formula (6).

The degree can be taken out of the sign of the logarithm

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! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.

Formula for moving to a new foundation

That rare case when the ODZ does not change during transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.

If we choose the number b as the new base c, we obtain an important special 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: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.

Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).

Table of formulas related to logarithms