# How to solve logarithmic equations?

To solve logarithmic equations, we have to use the laws of logarithms to rewrite the expressions in a more convenient way. After simplifying and reducing the logarithmic expressions, we will generally get one of two types of logarithmic equations. Depending on the type of equation obtained, we can obtain the answer simply by comparing the arguments of the logarithms or we can rewrite the logarithm in its exponential form to solve.

Here, we will look at a summary of the laws of logarithms. Also, we will look at the two types of logarithmic equations that we can obtain. Finally, we will learn how to solve logarithmic equations with examples.

##### ALGEBRA

Relevant for

Learning to solve logarithmic equations with examples.

See examples

##### ALGEBRA

Relevant for

Learning to solve logarithmic equations with examples.

See examples

## Types of logarithmic equations

Generally, we have two types of logarithmic equations. We need to recognize these two cases so that solving logarithmic equations becomes easier.

• The first type looks like this:

If we have only one logarithm on each side of the equation that has the same base, we can equalize the arguments of the logarithms and solve. In this case, the arguments are the algebraic expressions represented by P and Q.

• The second type looks like this:

If we have only one logarithm on only one side of the equation, then we can express it as an exponential expression and solve it that way.

## Summary of the laws of logarithms

### Product Law

When we have a logarithm of a product, we can write it as the sum of the individual logarithms of the factors:

### Quotient Law

When we have a logarithm of a quotient, we can write it as the logarithm of the numerator minus the logarithm of the denominator:

### Power Law

The logarithm of an exponential value can be rewritten as the exponent multiplied by the logarithm of the base (without the exponent):

### Law of Zero

The logarithm of 1 where the base different from zero and greater than zero is equal to zero:

### Law of identity

The logarithm of a value equal to the base is equal to 1. The base, b must be greater than zero, but cannot be equal to 1:

### Law of the logarithm of the exponent

The logarithm of an exponential value, where the base of the power equals the base of the logarithm, is equal to the exponent:

### Law of the exponent of a logarithm

Raising the logarithm of a number to its base equals the number:

## Learn to solve logarithmic equations with examples

### EXAMPLE 1

• Solve the logarithmic equation: $$\log_{3}(x+2)+\log_{3}(2)=\log_{3}(14)$$

Solution: We can use the product law to form a single logarithmic expression on the left hand side. Therefore, we have:

$latex \log_{3}(x+2)+\log_{3}(2)=\log_{3}(14)$

$latex \log_{3}[(x+2)2]=\log_{3}(14)$

Using the distributive property of multiplication, we have:

$latex \log_{3}(2x+4)=\log_{3}(14)$

We have a logarithm with the same base on each side, so we can form an equation with the arguments:

$latex 2x+4=14$

Now we have a linear equation, so we can easily solve:

$latex 2x+4=14$

$latex 2x=10$

$latex x=5$

### EXAMPLE 2

• Solve the logarithmic equation: $$\log_{2}(x)+\log_{2}(x+2)=\log_{2}(x+12)$$

Solution: Similar to the previous example, we can use the product law to form a single logarithm on the left side of the equation:

$$\log_{2}(x)+\log_{2}(x+2)=\log_{2}(x+12)$$

$latex \log_{2}[x(x+2)]=\log_{2}(x+12)$

Using the distributive property to distribute the x and obtain:

$latex \log_{2}({{x}^2}+2x)=\log_{2}(x+12)$

The logarithm on both sides of the equation has the same base, so we can eliminate it and form an equation with the arguments:

$latex {{x}^2}+2x=x+12$

In this case, we have a quadratic equation. We can move all the terms to one side of the equation and use factoring to solve:

$latex {{x}^2}+2x=x+12$

$latex {{x}^2}+2x-x-12=0$

$latex {{x}^2}+x-12=0$

$latex (x+4)(x-3)=0$

We make each factor equal to zero and solve:

⇒   $latex x=-4$

⇒   $latex x=3$

Therefore, we have two answers, $latex x=-4$ and $latex x=3$.

### EXAMPLE 3

• Solve the logarithmic equation: $$\log_{2}(x+4)-\log_{2}(3)=\log_{2}(x-2)-\log_{2}(5)$$

Solution: Here, we have a logarithm subtraction on each side of the equation. We can use the quotient law to obtain a single logarithm on each side:

$$\log_{2}(x+4)-\log_{2}(3)=\log_{2}(x-2)-\log_{2}(5)$$

$latex \log_{2}(\frac{x+4}{3})=\log_{2}(\frac{x-2}{5})$

We cannot reduce the expressions inside logarithms. However, since both logarithms have the same base, we can eliminate them and form an equation with the arguments:

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

We can simplify this expression by cross multiplying:

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

We apply the distributive property of multiplication to simplify:

$latex 5x+20=3x-6$

We have a linear equation, so we can easily solve:

$latex 5x+20=3x-6$

$latex 2x=-26$

$latex x=-13$

### EXAMPLE 4

• Solve the logarithmic equation: $latex \ln({{x}^2})+\frac{1}{2}\ln(4)=\ln({{x}^2}+16)$.

Solution: Here, we have natural logarithms, that is, logarithms with base e. These logarithms are denoted with ln. Although they are written slightly differently, all the laws of logarithms normally apply to natural logarithms.

We can use the power law to rewrite the logarithm that has a fraction in front:

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

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

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

Now, we can use the law of the logarithm of a product to simplify the left part of the equation:

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

We have a natural logarithm on each side, so we can eliminate it and write an equation with the arguments:

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

We can solve this quadratic equation easily:

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

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

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

Taking the square root of both sides, we have

$latex x=\sqrt{16}$

$latex x=\pm 4$

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

### EXAMPLE 5

• Solve the logarithmic equation: $latex \log(5x+40)=2$.

Solution: We see that we have the logarithm without any written base. When we have this, we assume that the logarithm has base 10. In fact, the logarithm with base 10 is called the common logarithm.

In this case, we only have a one-sided logarithm. We consider this as the second case mentioned above:

We are going to transform the equation from logarithmic to exponential form to solve. Therefore, the logarithm argument stays in place and we remove the logarithm. The number on the right becomes the exponent of the base of the logarithm, which is 10:

$latex \log(5x+40)=2$

$latex 5x+40={{10}^2}$

We simplify the exponent and solve the equation:

$latex 5x+40=100$

$latex 5x=60$

$latex x=12$

### EXAMPLE 6

• Solve the logarithmic equation: $latex \log_{2}(5x)-2=\log_{2}(2x-2)$.

Solution: We can move all logarithms to one side of the equation and the constant terms to the other side:

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

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

Now, we can use the quotient law to form a single logarithm on the left side:

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

Now, we are going to write the expression in its exponential form. The argument remains in the same place and we eliminate the logarithm. The 2 becomes the exponent of the base:

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

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

We can cross multiply to simplify:

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

We simplify with the distributive property of multiplication:

$latex 5x=8x-8$

We can easily solve the linear equation:

$latex 5x-8x=-8$

$latex -3x=-8$

$latex x=\frac{8}{3}$

Interested in learning more about logarithmic equations? Take a look at these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  