Lesson Objectives
• Demonstrate an understanding of exponents and logarithms
• Learn how to solve logarithmic equations with logarithms on each side
• Learn how to solve logarithmic equations with a logarithm equal to a number

## How to Solve Logarithmic Equations

Before we jump in and start solving logarithmic equations, let's look at some of the properties we will be using in this lesson.

### Properties for Solving Exponential and Logarithmic Equations

• b, x, and y are real numbers, b > 0, b ≠ 1
• If x = y, then bx = by
• If bx = by, then x = y
• If x = y and x > 0, y > 0, then logb(x) = logb(y)
• If x > 0, y > 0 and logb(x) = logb(y), then x = y

### Solving Logarithmic Equations

• Transform the equation so that a single logarithm appears on one side
• We can use the product rule or quotient rule for logarithms to accomplish this task
• Use one of the following rules to obtain a solution
• If x > 0, y > 0 and logb(x) = logb(y), then x = y
• If logb(x) = k, then x = bk
Let's look at a few examples.
Example 1: Solve each equation. $$\text{log}_{5}(30)=\text{log}_{5}(3x + 9)$$ To solve this equation, we use the following property:
If x > 0, y > 0 and logb(x) = logb(y), then x = y
Since we have the same base on each log, we can set the arguments equal to each other: $$3x + 9=30$$ Subtract 9 away from each side of the equation: $$3x=21$$ Divide each side by 3: $$x=7$$ Example 2: Solve each equation. $$9 - 3\text{log}_{8}(3x - 1)=6$$ For this scenario, we want to isolate the logarithm. Let's begin by subtracting 9 away from each side of the equation. $$-3\text{log}_{8}(3x - 1)=-3$$ Divide each side by -3: $$\text{log}_{8}(3x - 1)=1$$ To solve this equation, we use the following property:
If logb(x) = k, then x = bk
In other words, we write this in exponential form: $$3x - 1=8^1$$ $$3x - 1=8$$ Add 1 to each side of the equation: $$3x=9$$ Divide each side by 3: $$x=3$$

### Extraneous Solutions with Logarithmic Equations

In our previous examples, we did not show a check to prove our answers were correct. When working with logarithmic equations, you may end up with extraneous solutions or solutions that do not work in the original equation. This problem may come up when using the properties of logarithms to condense. We know the argument of a logarithm is undefined for non-positive numbers (0 or negative). When multiplying or dividing with two negative numbers, we create a positive number. Therefore, something that is undefined in the original equation may work in the modified version, creating extraneous solutions or solutions that do not work in the original equation. Therefore, it is advised that you always check your solution(s) in the original problem.
Example 3: Solve each equation. $$\text{log}_{6}(x + 9) + \text{log}_6(x + 10)=1$$ Here, we will use the product rule for logarithms to condense the left side: $$\text{log}_{6}[(x + 9)(x + 10)]=1$$ Simplify: $$\text{log}_{6}(x^2 + 19x + 90)=1$$ To solve this equation, we use the following property:
If logb(x) = k, then x = bk
In other words, we write this in exponential form: $$6^1=x^2 + 19x + 90$$ $$x^2 + 19x + 90=6$$ Subtract 6 away from each side of the equation: $$x^2 + 19x + 84=0$$ Factor the left side: $$(x + 7)(x + 12)=0$$ Use the zero-product property to solve for x: $$x + 7=0$$ $$x=-7$$ $$x + 12=0$$ $$x=-12$$ Let's check our solutions and see if they work in the original equation. $$\text{log}_{6}(x + 9) + \text{log}_6(x + 10)=1$$ Plug in a -7 for each x: $$\text{log}_{6}((-7) + 9) + \text{log}_6((-7) + 10)=1$$ $$\text{log}_{6}(2) + \text{log}_6(3)=1$$ $$\text{log}_{6}(2 \cdot 3)=1$$ $$\text{log}_{6}(6)=1$$ $$1=1{\color{green}✓}$$ Plug in a -12 for each x: $$\text{log}_{6}((-12) + 9) + \text{log}_6((-12) + 10)=1$$ $$\text{log}_{6}(-3) + \text{log}_6(-2)=1$$ Since logarithms of negative numbers are undefined, we would state that x = -12 is an extraneous solution. It does not work in the original equation. Therefore, our only valid solution here is x = -7.

### Common Errors When Checking Solutions for Logarithmic Equations

A common error sometimes occurs when students check in the transformed version of the equation. $$\text{log}_{6}(x^2 + 19x + 90)=1$$ Notice that if we started with this equation, both solutions of x = -7 and x = -12 will work. This happens due to the fact that the product of two negative numbers gives us a positive. So the multiplication that occurs when condensing is causing the issue.
This issue is obvious if you look at the graph for each. Let's start with the original equation. $$\text{log}_{6}(x + 9) + \text{log}_6(x + 10)=1$$ We will subtract 1 away from each side: $$\text{log}_{6}(x + 9) + \text{log}_6(x + 10) -1=0$$ Now, let's graph: $$y=\text{log}_{6}(x + 9) + \text{log}_6(x + 10) -1$$ The solution is the x-value for which the y-value is zero, in other words, where is the x-intercept? Now, let's graph the transformed version of the equation. $$\text{log}_{6}(x^2 + 19x + 90)=1$$ We will subtract 1 away from each side: $$\text{log}_{6}(x^2 + 19x + 90) - 1=0$$ Now, let's graph: $$y=\text{log}_{6}(x^2 + 19x + 90) - 1$$ The solutions are the x-values for which the y-value is zero, in other words, where are the x-intercepts?

#### Skills Check:

Example #1

Solve each equation. $$\text{log}_{9}(x - 8) + \text{log}_{9}(10)=\text{log}_{9}(80)$$

A
$$x=\frac{1}{2}$$
B
$$x=4$$
C
$$x=\frac{41}{4}$$
D
$$x=18$$
E
$$x=16$$

Example #2

Solve each equation. $$\text{log}_{6}(-4x) - \text{log}_{6}(2)=3$$

A
$$x=-\frac{3}{4}$$
B
$$x=36$$
C
$$x=-\frac{17}{2}$$
D
$$x=-\frac{15}{4}$$
E
$$x=-108$$

Example #3

Solve each equation. $$\text{log}_{5}(4x) - \text{log}_{5}(6)=1$$

A
$$x=-1$$
B
$$x=2$$
C
$$x=\frac{15}{2}$$
D
$$x=\frac{4}{5}$$
E
$$x=\frac{5}{3}$$

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