# Logarithmic Equations – Examples and Practice Problems

Logarithmic equation exercises can be solved using the laws of logarithms. With the laws of logarithms, we can rewrite logarithmic expressions to get more convenient expressions. Depending on the problem, we can end up with two types of logarithmic equations with which we will have to use different methods to get the answer.

Here, we will review the process used to solve logarithmic equation exercises. We will look at a summary of the two methods that we can apply to obtain the answer. In addition, we will look at several examples with answers to fully master the topic of logarithmic equations.

##### ALGEBRA

Relevant for

Learning to solve logarithmic equations with examples.

See examples

##### ALGEBRA

Relevant for

Learning to solve logarithmic equations with examples.

See examples

## Summary of logarithmic equations

Logarithmic equations can be solved using the laws of logarithms. These laws allow us to rewrite logarithms and form more convenient expressions. If you need to review the laws of logarithms, you can look at this article: Laws of Logarithms.

The goal is to reduce to the logarithmic equation until you get a single logarithm on each side or a single logarithm on one side. Based on this, we can distinguish two types of logarithmic equations. We have to recognize these two types to facilitate solving the equations.

### Types of logarithmic equations

Generally, after applying the laws of logarithms to reduce the equation, we can end up with one of two types of logarithmic equations:

• The first type looks like this:

In cases where we end up with only one logarithm on each side of the equation, we can eliminate the logarithms if they have the same base and we can form an equation with the arguments. For example, in the expression above, the arguments are the algebraic expressions represented by P and Q.

• The second type looks like this:

In cases where we end up with a single logarithm on only one side of the equation, we can write the logarithm as an exponential expression and solve it that way.

## Logarithmic equations – Examples with answers

The following logarithmic equation examples use the laws of logarithms and both methods detailed above. Each example has its respective answer, but it is recommended that you try to solve the exercises yourself before looking at the solution.

### EXAMPLE 1

What is the result of $latex \log_{5}(x+1)+\log_{5}(3)=\log_{5}(15)$?

We see that we have a sum of logarithms with the same base on the right-hand side, so we can use the product law to combine them. This is the product law in case you don’t remember it:

Therefore, we have:

$latex \log_{5}(x+1)+\log_{5}(3)=\log_{5}(15)$

$latex \log_{5}[(x+1)3]=\log_{5}(15)$

We can expand the multiplication to get:

$latex \log_{5}(3x+3)=\log_{5}(15)$

The logarithms have the same base, so we can eliminate them and form an equation with the arguments:

$latex 3x+3=15$

The linear equation can be easily solved:

$latex 3x+3=15$

$latex 3x=12$

$latex x=4$

### EXAMPLE 2

Solve the equation $$\log_{4}(2x+2)+\log_{4}(2)=\log_{4}(x+1)+\log_{4}(3)$$

In this case, we have a sum of logarithms on each side of the equation. Therefore, we are going to use the law of the product on both sides to get:

$$\log_{4}(2x+2)+\log_{4}(2)=\log_{4}(x+1)+\log_{4}(3)$$

$latex \log_{4}[(2x+2)2]=\log_{4}[(x+1)3]$

We can expand the multiplication on both sides to get:

$latex \log_{4}(4x+4)=\log_{4}(3x+3)$

Now, we eliminate the logarithms and form an equation with the arguments:

$latex 4x+4=3x+3$

The linear equation can be easily solved:

$latex 4x+4=3x+3$

$latex x=-1$

### EXAMPLE 3

Solve the equation $$\log_{7}(x)+\log_{7}(x+5)=\log_{7}(2x+10)$$

We can use the product law on the left side:

$$\log_{7}(x)+\log_{7}(x+5)=\log_{7}(2x+10)$$

$latex \log_{7}[x(x+5)]=\log_{7}(2x+10)$

Now, we can distribute the x to get:

$latex \log_{7}({{x}^2}+5x)=\log_{7}(2x+10)$

We can eliminate the logarithm on both sides since they have the same base and we can form an equation with the arguments:

$latex {{x}^2}+5x=2x+10$

We obtained a quadratic equation. We can solve this equation by placing all the terms on one side and factoring the equation:

$latex {{x}^2}+5x=2x+10$

$latex {{x}^2}+5x-2x-10=0$

$latex {{x}^2}+3x-10=0$

$latex (x+5)(x-2)=0$

We solve for each factor:

⇒   $latex x=-5$

⇒   $latex x=2$

Therefore, we have two answers, $latex x=-5$ and $latex x=2$.

### EXAMPLE 4

What is the value of x in $$\log_{3}(x+3)-\log_{3}(2)=\log_{3}(x-1)-\log_{3}(7)$$

In this case, we have log subtractions on both sides of the equation, so we can apply the law of the logarithm quotient. In case you don’t remember, the following is the quotient law:

Therefore, applying this law to both sides, we have:

$$\log_{3}(x+3)-\log_{3}(2)=\log_{3}(x-1)-\log_{3}(7)$$

$latex \log_{3}(\frac{x+3}{2})=\log_{3}(\frac{x-1}{7})$

Expressions within logarithms can no longer be simplified. However, we can eliminate the logarithms since they both have the same base:

$latex \frac{x+3}{2}=\frac{x-1}{7}$

We can cross multiply to simplify:

$latex 7(x+3)=2(x-1)$

We multiply using the distributive property:

$latex 7x+21=2x-2$

We solve the linear equation:

$latex 7x+21=2x-2$

$latex 5x=-23$

$latex x=-\frac{23}{5}$

### EXAMPLE 5

What is the result of $latex \log({{x}^2})+\frac{1}{2}\log(4)=\log({{x}^2}+16)$?

We can see that the logarithms in this equation do not have a base. When we have logarithms without a base, we assume that the base is 10. Logarithms with base 10 are called common logarithms.

In this equation, we can start by using the power law to rewrite the logarithm that has a fraction in front of it.

Therefore, we have:

$latex \log({{x}^2})+\frac{1}{2}\log(4)=\log({{x}^2}+16)$

$latex \log({{x}^2})+\log({{4}^{\frac{1}{2}}})=\log({{x}^2}+16)$

$latex \log({{x}^2})+\log(2)=\log({{x}^2}+16)$

We simplify the left side using the product law:

$latex \log(2{{x}^2})=\log({{x}^2}+16)$

We can eliminate the logarithm on each side since it has the same base:

$latex 2{{x}^2}={{x}^2}+16$

We can solve for x to solve the quadratic equation:

$latex 2{{x}^2}={{x}^2}+16$

$latex 2{{x}^2}-{{x}^2}=16$

$latex {{x}^2}=16$

Now, we take the square root of both sides:

$latex x=\sqrt{16}$

$latex x=\pm 4$

Therefore, we have two answers, $latex x=4$ and $latex x=-4$.

### EXAMPLE 6

What is the value of x in $latex \log(4x+60)=2$?

Here we also don’t have a base in the logarithm, so we know that it is a common logarithm and that its base is 10.

In this equation, we have a one-sided logarithm. This is an equation of the second case mentioned above:

We can solve this equation by writing it in exponential form. Therefore, we remove the logarithm from the left side and write its argument. On the right-hand side, 2 is the exponent and 10 is the base (the base of the logarithm):

$latex \log(4x+60)=2$

$latex 4x+60={{10}^2}$

We apply the exponent and solve the linear equation:

$latex 4x+60=100$

$latex 4x=40$

$latex x=10$

### EXAMPLE 7

Find the value of x in the equation $latex \log_{2}(3x)-2=\log_{2}(2x-5)$.

We have to move the logarithms to one side of the equation and the constant terms to the other side:

$latex \log_{2}(3x)-2=\log_{2}(2x-5)$

$latex \log_{2}(3x)-\log_{2}(2x-5)=2$

Now, we simplify the left part using the quotient law:

$latex \log_{2}(\frac{3x}{2x-5})=2$

To solve, we have to write the equation in its exponential form. The argument remains in the same place and we eliminate the logarithm. We raise the 2 (the base of the logarithm) to the exponent 2:

$latex \frac{3x}{2x-5}={{2}^2}$

$latex \frac{3x}{2x-5}=4$

We simplify by multiplying crosswise:

$latex 3x=4(2x-5)$

We distribute the multiplication:

$latex 3x=8x-20$

We solve the linear equation:

$latex 3x=8x-20$

$latex -5x=-20$

$latex x=\frac{-20}{-5}$

$latex x=4$

## Simplifying algebraic expressions – Practice problems

Put into practice what you have learned about logarithmic equations with the following problems. Solve the problems and choose your answer. Check your answer to verify that you selected the correct one.