Equations containing variables in logarithmic expressions are called logarithmic equations (sometimes shortened as “log equations”). Solving logarithmic equations can be easy and entertaining if you are aware of the principal methods and different scenarios. Here, we’ll provide a comprehensive guide on the most efficient methods to solve log equations.

Type 1 Logarithmic Equations

The simplest logarithmic equations are equations of the form

log_b x = a

where the base of the logarithm, b, is a positive number, b ≠ 1. For any real value of the variable x, this equation has a single solution:

x = b^a

Example 1

For example, the log equation

log_2 x = -, 3

only has one solution:

x = 2^{ - 3 } = \dfrac{ 1 }{ 8 }

Example 2

Consider logarithmic equations of the form

log_x b = a

This can be simplified by using the change of base formula for logarithms:

log_x b = \dfrac{ log_b b }{ log_b x } = \dfrac{ 1 }{ log_b x }

This results in

log_b x = a^{ - 1 }

For example, the log equation

log_x 10 = 5

can be written in the following form:

log x = \dfrac{ 1 }{ 5 }

This has the solution

x = 10^{ 1/5 }

Type 2 Logarithmic Equations

Consider a slightly more complicated logarithmic equation:

log_b f ( x ) = a

Here, again, the base of the logarithm, b, is a positive number, b ≠ 1, and f(x) is some elementary algebraic function. This equation can be solved by introducing a new variable, t = f(x), defined as

log_b t = a

Thus, we can write the following general solution for this equation:

t = f ( x ) = b^a

Let’s illustrate this idea with examples.

Example 1

Say we want to solve the following logarithmic equation:

log_3( x^2 + x + 3 ) = 2

We substitute the following variable:

t = x^2 + x + 3

This turns the equation into a standard form:

log_3 t = 2

This has the solution

t = 3^2 = 9

Thus, we have a quadratic equation for the variable x:

x^2 + x + 3 = 9

We can easily determine the roots of this quadratic equation, which are also the solutions of the original logarithmic equation:

x_1=2 ,, \quad x_2 = -3

Example 2

Consider the following log equation:

log_{ ( x^2 + 1 ) } 10 = 1

This can be solved after applying the change of base formula:

log_{ ( x^2 + 1 ) } 10 = \dfrac{ log 10 }{ log ( x^2 + 1 ) } = \dfrac{ 1 }{ log ( x^2 + 1 ) }

We must also substitute a new variable:

t = x^2 +1

For the variable t, we have

log t = 1

This results in t = 10. Thus, we obtain

x^2 + 1 = 10

Both roots of this algebraic equation are solutions to the original logarithmic equation:

x_1 = 3 ,, \quad x_2 = -, 3

Example 3

This method also works if f(x) is a logarithmic function by itself. For example, consider the equation

ln( log x ) = 0

As usual, we can introduce a new variable t, according to the following equation:

t = log x

Then, t is a solution to the simple log equation

ln t = 0

which can be solved easily:

t = e^0 = 1

The variable x also becomes the solution to a simple log equation:

log x = 1

Finally, we can find the solution:

x = 10^1 = 10

Type 3 Logarithmic Equations

Next, we will investigate how to solve log equations of the form

log_b f ( x ) = log_b h ( x )

where f(x) and h(x) are some elementary algebraic functions, and b is a positive number, b ≠ 1. This log equation is equivalent to the algebraic equation

f ( x ) = h ( x )

We should also remember that the domain of any logarithmic function is nonnegative, real numbers. Thus, among all the solutions to the equation f(x) = h(x), we should only select solutions that satisfy one of the following conditions:

f ( x ) > 0 \quad \quad \text{ or} \quad \quad h ( x ) > 0

This guarantees that the logarithmic functions are well-defined.

Example 1

As an example, let us consider the equation

3 log x = log 2 x

Since

3 log x = log x^3

we have

x^3 = 2 x

Among all the roots of this equation, which can be defined as

x_1 = 0 ,, \quad \quad x_2 = - , \sqrt{ 2 } ,, \quad \quad x_3 = \sqrt{ 2 }

Only those that satisfy the condition

\quad \quad x > 0

can be solutions of the original logarithmic equation. Hence, roots x₁ and x₂ should be rejected, and the only solution is

x = \sqrt{ 2 }

Example 2

ln (2-x) = - ,ln x

If we take into account that

ln \dfrac{ 1 }{ x } = -, ln x

We can rewrite this log equation as

ln (2-x) = ln \dfrac{ 1 }{ x }

Applying the method described for this type of logarithmic equation, we obtain

2 - x = \dfrac{ 1 }{ x } ,, \quad \quad x > 0

This equation is equivalent to

( x - 1)^2 = 0

This has a single solution:

x = 1

Obviously, this solution satisfies the condition x > 0.

Example 3

The method we are now investigating only requires a slight modification if the logarithmic expressions on both sides of the log equation have different bases:

log_a f ( x ) = log_b h ( x )

where both a and b are positive numbers, a ≠ 1 and b ≠1. We can rewrite the left-hand side of this equation in the form

log_a f ( x ) = \dfrac{ log_b f ( x ) }{ log_b a }

Thus, we have

log_b f ( x ) = log_b a log_b h ( x ) = log_b \left( h^{ log_b a } \right)

Then, the log equation in question is again equivalent to the following algebraic equation:

f = h^{ log_b a }

Pay attention to the fact that both of the following conditions must be satisfied in this case:

f ( x ) > 0 \quad \quad \text{ and} \quad \quad h ( x ) > 0

To illustrate, consider the log equation

log_2 x = log_4 ( x + 2 )

Using the change of base formula, we have

log_4 ( x + 2 ) = \dfrac{ log_2 ( x + 2 ) }{ log_2 4 } = \dfrac{ 1 }{ 2 } log_2 ( x + 2 )

Thus, we can write

log_2 x^2 = log_2 ( x + 2 )

This results in the following algebraic equation:

x^2 = x + 2

Further, remember that both initial logarithmic functions are defined only in the region

x > 0

Thus, between the two roots of the quadratic equation

x_1 = -,1 ,, \quad \quad x_2 = 2

only x = 2 is a solution to the original log equation. 

Type 4 Logarithmic Equations

A more challenging class of logarithmic equations are equations of the form

log_b f_1(x) + log_b f_2(x) + \ldots + log_b f_n(x) = log_b h_1(x) + log_b h_2(x) + \ldots + log_b h_m(x)

where b is a positive number, b ≠ 1, and

f_i ( x ),, \quad i = 1, 2, \ldots , n

h_j ( x ),, \quad j = 1, 2, \ldots , m

These are some algebraic functions (some of them can be constant numbers).

Solving logarithmic equations of this type is equivalent to solving the following system of algebraic equations:

f_1(x) f_2(x) \ldots f_n(x) = h_1(x) h_2(x) \ldots h_m(x)

f_i ( x ) > 0,, \quad i = 1, 2, \ldots , n

h_j ( x ) > 0,, \quad j = 1, 2, \ldots , m

Example 1

For example, consider the following log equation:

ln 2 x + ln ( x - 1) = ln 4

As we have just discussed, to solve this logarithmic equation, we have to solve

2 x ( x - 1 ) = 4

x > 0 ,, \quad \quad x > 1

Among the two roots of the quadratic equation,

x_1 = -, 1 ,, \quad \quad x_2 = 2

only the second one satisfies the above inequalities. Thus, the only solution of the given logarithmic equation is

x = 2

Example 2

log 2 x + log ( x + 3 ) = log 2 + log ( 6 x - 2 )

Based on our general consideration that solving this logarithmic equation is equivalent to solving a mixed system of algebraic equations,

2 x ( x + 3 ) = 2 ( 6 x - 2 )

2 x > 0

x + 3 > 0

6 x - 2 > 0

These, in turn, are equivalent to the system

x^2 - 3 x + 2 = 0

x > \dfrac { 1 } { 3 }

Both roots of the quadratic equation,

x_1 = 1 \quad \text{ and } \quad x_2 = 2

satisfy the inequality x > 1/3. Thus, both roots are solutions of the logarithmic equation.

Example 3

Consider the following equation:

log ( 2 x^2 + 21 x + 9 ) - log ( 2 x + 1 ) = 1

This belongs to the same class of logarithmic equations. It is not difficult to observe that it can be written in the form

log ( 2 x^2 + 21 x + 9 ) = log 10 + log ( 2 x + 1 )

This equation is equivalent to a mixed system of algebraic equations:

2 x^2 + 21 x + 9 = 10 ,( 2 x + 1 )

2 x + 1 > 0

2 x^2 + 21 x + 9 > 0

The quadratic equation can be written in the form

2 x^2 + x - 1 = 0

This has two roots:

x_1 = - ,1 ,, \quad \quad x_2 = \dfrac{ 1 }{ 2 }

We can easily verify that the first root, x₁ = – 1, does not satisfy the inequality 2 x + 1 > 0, while the second root, x₂ = 1/2, satisfies both inequalities. Thus,

x = \dfrac{ 1 }{ 2 }

This is the only solution to the logarithmic equation in question.

Example 4

If we encounter a problem with fractions of logarithms, most likely, the log equation can be transformed into the same standard form. For example, if we need to solve

\dfrac{ log_6 15x }{ log_6 5x } = 2

First, we note that

x > 0

This is the range for which the variable x is valid. Next, rewrite the equation in the form

log_6 15x = 2 log_6 5x = log_6 25 x^2

Thus, the logarithmic equation under investigation becomes an algebraic equation:

15 x = 25 x^2

This can be solved easily:

x_1 = 0 ,, \quad \quad x_2 = \dfrac{ 3 }{ 5 }

Finally, the only root satisfying the condition x > 0 is

x = \dfrac{ 3 }{ 5 }

Type 5 Logarithmic Equations

Finally, we will learn how to solve logarithmic equations of the form

F[ h ( x ) ] = 0

Here, h(x) is some logarithmic function and F(u) is an elementary algebraic function. In this case, we can introduce a new variable t = h(x) and solve:

F ( t ) = 0

Let

t = t_1 ,, t_2 ,, \ldots ,, t_n

be the n real numbers that are solutions to the algebraic equation F(t) = 0. Then, to solve the original logarithmic equation, we have to find solutions to the following system of n algebraic equations:

h ( x ) = t_1

h ( x ) = t_2

\ldots

h ( x ) = t_n

Example 1

log \sqrt{ x } = \sqrt{ log x }

Using the fact that

log \sqrt{ x } = \dfrac{ 1 }{ 2 }, log x

we can write this log equation in the following form:

\dfrac{ 1 }{ 2 }, log x = \sqrt{ log x }

For the auxiliary variable

t = log x

we obtain a simple algebraic equation:

\dfrac{ t }{ 2 } = \sqrt{ t }

This has the following roots:

t_1 = 0 ,, \quad \quad t_2 = 4

Thus, we have to solve two logarithmic equations:

log x = 0

log x = 4

This is quite an easy task. The corresponding solutions are

x_1 = 10^0 = 1 ,, \quad \quad x_2 = 10^4 = 10000

These are both in the range of validity for the logarithmic function, x > 0.

Example 2

Consider another logarithmic equation:

log x^3 ,( log x -5 ) + 18 = 0

Since

log x^3 = 3 log x

we can rewrite this equation in the form

3 log x ,(log x -5 ) + 18 = 0

Now, we introduce a new variable:

t = log x

This is the solution of the equation

3 t ( t - 5 ) + 18 = 0

After some simple algebraic transformations, we obtain

t^2 -5 t + 6 = 0

t_1 = 2 ,, \quad \quad t_2 = 3

Thus, we have to solve the following log equations:

log x = 2

log x = 3

The obvious solutions are

x_1 = 10^2 = 100 ,, \quad \quad x_2 = 10^3 = 1000

Example 3

log_x 3 log_{ 3x } 3 = log_{ 9x } 3

Although this log equation seems quite different from those we have considered so far, we will now show how it can be solved using the same methods.

First, since the base of a logarithm can only be a positive number not equal to unity, the domain of validity for the variable x can be expressed as

x > 0 ,, \quad \quad x \neq 1,, \dfrac{ 1 }{ 3 },, \dfrac{ 1 }{ 9 }

Now, we can use a change of base formula for logarithms to express log_x 3 in terms of log₃ x:

log_x 3 = \dfrac{ log_3 3 }{ log_3 x } = \dfrac{ 1 }{ log_3 x }

Analogously, we can write

log_{ 3 x } 3 = \dfrac{ 1 }{ log_3 3 x } = \dfrac{ 1 }{ 1 + log_3 x }

log_{ 9 x } 3 = \dfrac{ 1 }{ log_3 9 x } = \dfrac{ 1 }{ 2 + log_3 x }

The original logarithmic equation now transforms into an equation of standard form:

log_3 x ,( 1 + log_3 x ) = 2 + log_3 x

This can be solved by the following substitution:

t = log_3 x

The equation in terms of the variable t takes the form:

t ,( 1 + t ) = 2 + t

We can easily find the corresponding roots:

t_1 = \sqrt{ 2 },, \quad \quad t_2 = -,\sqrt{ 2 }

Thus, we have two log equations:

log_3 x = \sqrt{ 2 }

log_3 x = -,\sqrt{ 2 }

These are extremely easy to solve, yielding the following results:

x_1 = 3^{ \sqrt{ 2 } } ,, \quad \quad x_2 = 3^{ - \sqrt{ 2 } } = \dfrac{ 1 }{ 3^{ \sqrt{ 2 } } }

Both solutions are in the valid range for the variable x.

Hopefully, this review article has improved your understanding of how to solve logarithmic equations in algebra. If you work through the wide range of examples presented here, you will be prepared to solve logarithmic equations of any difficulty. Best of luck!

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