Math Lesson 13.2.2 - Logarithmic Equations

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Welcome to our Math lesson on Logarithmic Equations, this is the second lesson of our suite of math lessons covering the topic of Exponential and Logarithmic Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Logarithmic Equations

Logarithmic equations are those equations that contain the variable in the argument of a logarithm. For example,

are all logarithmic equations, as in all cases the variable is in the argument of a logarithm.

There are three main types of logarithmic equations:

1. Equations with a single logarithmic term;
2. Equations with more than one logarithmic term but with the same base; and
3. Equations with more than one logarithmic term with different bases. Let's have a look at each of them.

1. Equations with a single logarithmic term

In these equations, we send the variable out of the logarithm by using the logarithmic to exponential conversion. The general form of such equations is

where x is the variable. Let's try to understand more through a pair of examples.

Example 6

Solve the following logarithmic equations.

1. log₃ x = 2
2. log₅ (2x - 1) = 3
3. log₂ (1 - 5x) = 4

Solution 6

We must write all equations in the exponential form. Thus,

1. log₃ x = 2
   x = 3²
   x = 9
2. log₅ (2x - 1) = 3
   2x - 1 = 5³
   2x - 1 = 125
   2x = 125 + 1
   2x = 126
   x = 126/2
   x = 63
3. log₂ (1 - 5x) = 4
   1 - 5x = 2⁴
   1 - 5x = 16
   -5x = 16 - 1
   -5x = 15
   x = 15/-5
   x = -3

2. Equations with more than one logarithmic term but with the same base

In these equations we only have to focus on the arguments to solve them. This means we can disregard the bases during the solution and solely consider the arguments. However, this is completed only after having expressed the equation to a single logarithmic term on either side. Look at the examples below.

Example 7

Solve the following logarithmic equations.

1. log₅ (1 - 2x) = log₅ 13
2. log₂ (3 - 4x) = log₂ (x-6)
3. log₃ (3x - 1) = 2 log₃ 9

Solution 7

1. Since the base is the same in both sides, so the arguments must be as well. Thus,
   log₅ (1 - 2x) = log₅ 13
   1 - 2x = 13
   -2x = 13 - 1
   -2x = 12
   x = 12/-2
   x = -6
2. Again, since the base is the same in both sides, so must the arguments be as well. Thus,
   log₂ (3 - 4x) = log₂ (x - 6)
   3 - 4x = x - 6
   3 + 6 = x + 4x
   9 = 5x
   x = 9/5
3. First, we have to write both sides as logarithms not multiplied by constants. For this, we have to apply the logarithmic property
   m ∙ log_a b = log_a b^m
   Hence, we write
   log₃ (3x - 1) = 2 log₃ 9
   log₃ (3x - 1) = log₃ 9²
   Now, we can focus only on the argument given that the rest of elements are equal. Thus,
   3x - 1 = 9²
   3x - 1 = 81
   3x = 81 + 1
   3x = 80
   x = 80/3

3. Equations with more than one logarithmic term with different bases

In these equations, the first thing to do is to attempt to write the logarithms with the same base. If this is possible, the equation is solved by using the second method described above; otherwise, the result is usually obtained by applying the logarithmic property

Let's explain this point through a pair of examples.

Example 8

Solve the following equations for x.

1. 1/2 log₂ (3x - 1) = log₄ (2x)
2. 2 log₅ (1 - 4x) = 4 log₂₅ (3x - 2)
3. log₂ (2x - 3) = log₆ 10

Solution 8

1. First, we turn both logarithms at the same base (here, 2). Thus,
   1/2 log₂ (3x - 1) = log₄ (2x)
   1/2 log₂ (3x - 1) = log_2² (2x)
   Now, we apply the property of logarithms
   log_aⁿ b = 1/n log_a b
   Thus, we obtain
   1/2 log₂ (3x - 1) = 1/2 log₂ (2x)
   log₂ (3x - 1) = log₂ (2x)
   3x - 1 = 2x
   3x - 2x = 1
   x = 1
2. Again, we first turn both logarithms at the same base (here, 2). Thus,
   2 log₅ (1 - 4x) = 4 log₂₅ (3x - 2)
   2 log₅ (1 - 4x) = 4 log_5² (3x - 2)
   2 log₅ (1 - 4x) = 1/2 ∙ 4 log₅ (3x - 2)
   2 log₅ (1 - 4x) = 2 log₅ (3x - 2)
   log₅ (1 - 4x) = log₅ (3x - 2)
   1 - 4x = 3x - 2
   1 + 2 = 3x + 4x
   3 = 7x
   x = 3/7
3. This time it is impossible to write the logarithmic terms at the same base. Thus, we have
   log₂ (2x - 3) = log₆ 10
   log₂ (2x - 3) = log 10/log 6
   log₂ (2x - 3) = 1/0.778
   log₂ (2x - 3) = 1.285
   2x - 3 = 2^1.285
   2x - 3 = 2.44
   2x = 2.44 + 3
   2x = 5.44
   x = 5.44/2
   x = 2.72

Obviously, there are more challenging exponential and logarithmic equations available in various textbooks; here we have shown only a few of them just for illustration purposes.

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