## How to Solve Logarithmic Equations

### How do we solve logarithmic equations?

To solve logarithmic equations, we must use the rules of logarithms to rewrite the expressions in a way that is easier to understand. After rewriting the logarithmic expressions, we usually get one of two types of logarithmic equations. Depending on the equation type, we can get the answer by comparing the logarithms' arguments or by writing the logarithm in its exponential form.

### Types of logarithm equations

Logarithmic equations are usually one of two types. We need to know about these two types so that it's easier to solve logarithmic equations.

• First type: $$log_b \ m \ = \ log_b \ n$$
If there is only one logarithm on each side of the equation and they both have the same base, we can make the logarithms' arguments equal and solve. In this case, the arguments are $$m$$ and $$n$$
$$log_b \ m \ = \ log_b \ n \ ⇒ \ m \ = \ n$$
• Second type: $$log_b \ m \ = \ n$$
If there is only one logarithm on one side of the equation, we can write it as an exponential equation and solve it that way.

While solving logarithmic equations, we need to use some of the logarithms rules. In case you've forgotten them, check this article (What are the Properties of Logarithms) before solving an example.

### Example1

Solve for $$x$$: $$log_3 \ 5 \ + \ log_3 \ (x \ + \ 2) \ = \ log_3 \ -5$$

Solution

We see that the left-hand side is a sum of logarithms with the same base, so we can use the product rule to add them together. If you can't remember the product rule, here it is: $$log_b \ (mn) \ = \ log_b \ m \ + \ log_b \ n$$
So, here's what we have: $$log_3 \ 5 \ + \ log_3 \ (x \ + \ 2) \ = \ log_3 \ -5 \ ⇒ \ log_3 \ 5(x \ + \ 2) \ = \ log_3 \ -5 \ ⇒ \ log_3 \ (5x \ + \ 10) \ = \ log_3 \ -5$$

Since the logarithms have the same base, we can get rid of them and use the arguments to make an equation: $$5x \ + \ 10 \ = \ -5$$

It's easy to solve the linear equation: $$5x \ + \ 10 \ = \ -5 \ ⇒ \ 5x \ = \ -15 \ ⇒ \ x \ = \ -3$$

### Example2

Solve for $$x$$: $$log_2 \ 4 \ + \ log_2 \ 2x \ = \ 5$$

Solution

In this equation, we have a logarithm on only one side. This is the second type of logarithmic equation described above.

This equation can be solved by writing it as an exponential equation. We see that the left-hand side is a sum of logarithms with the same base, so we can use the product rule to add them together: $$log_2 \ 4 \ + \ log_2 \ 2x \ = \ 5 \ ⇒ \ log_2 \ 8x \ = \ 5$$
Now, we take the logarithm off the left side of the equation and write its argument. On the right, $$5$$ is the exponent, and $$2$$ is the base:

$$log_2 \ 8x \ = \ 5 \ ⇒ \ 8x \ = \ 2^5 \ ⇒ \ 8x \ = \ 32 \ ⇒ \ x \ = \ 4$$

### Exercises for Solving Logarithmic Equations

1) Solve: $$log \ 8 \ + \ 5x \ = \ 0$$

2) Solve: $$log_2 \ 5 \ + \ log_2 \ 2x \ = \ 5$$

3) Solve: $$log_3 \ 4 \ + \ log_3 \ -3x \ = \ 6$$

4) Solve: $$log_5 \ x \ + \ log_5 \ 5x \ = \ 3$$

5) Solve: $$log_4 \ 2x \ + \ log_4 \ 7 \ = \ 4$$

6) Solve: $$5 \ log_2 \ -2x \ = \ 35$$

7) Solve: $$5 \ log \ 12 \ - \ 25x \ = \ 0$$

8) Solve: $$log_7 \ (-2x \ + \ 3) \ - \ 7 \ = \ 5$$

9) Solve: $$log_5 \ (4x \ + \ 5) \ = \ 3$$

10) Solve: $$log_9 \ (2x \ + \ 1) \ + \ log_9 \ 5x \ = \ 2$$

1) Solve: $$log \ 8 \ + \ 5x \ = \ 0$$

$$\color{red}{log \ 8 \ + \ 5x \ = \ 0 \ ⇒ \ log \ 8 \ = \ -5x \ ⇒ \ x \ = \ -\frac{log \ 8}{5} \ ≈ \ -0.1806}$$

2) Solve: $$log_2 \ 5 \ + \ log_2 \ 2x \ = \ 5$$

$$\color{red}{log_2 \ 5 \ + \ log_2 \ 2x \ = \ 5 \ ⇒ \ log_2 \ 5 \times 2x \ = \ 5 \ ⇒ \ 10x \ = \ 2^5 \ ⇒ \ x \ = \ 3.2}$$

3) Solve: $$log_3 \ 4 \ + \ log_3 \ -3x \ = \ 6$$

$$\color{red}{log_3 \ 4 \ + \ log_3 \ -3x \ = \ 6 \ ⇒ \ log_3 \ 4 \times (-3x) \ = \ 6 \ ⇒ \ -12x \ = \ 3^6 \ ⇒ \ x \ = \ -60.75}$$

4) Solve: $$log_5 \ x \ + \ log_5 \ 5x \ = \ 3$$

$$\color{red}{log_5 \ x \ + \ log_5 \ 5x \ = \ 3 \ ⇒ \ log_5 \ x \times 5x \ = \ 3 \ ⇒ \ 5x^2 \ = \ 5^3 \ ⇒ \ x^2 \ = \ 25 \ ⇒ \ x \ = \ 5}$$

5) Solve: $$log_4 \ 2x \ + \ log_4 \ 7 \ = \ 4$$

$$\color{red}{log_4 \ 2x \ + \ log_4 \ 7 \ = \ 4 \ ⇒ \ log_4 \ 2x \times 7 \ = \ 4 \ ⇒ \ 14x \ = \ 4^4 \ ⇒ \ x \ = \ 18.285}$$

6) Solve: $$5 \ log_2 \ -2x \ = \ 35$$

$$\color{red}{5 \ log_2 \ -2x \ = \ 35 \ ⇒ \ log_2 \ -2x \ = \ 7 \ ⇒ \ -2x \ = \ 2^7 \ ⇒ \ x \ = \ -64}$$

7) Solve: $$5 \ log \ 12 \ - \ 25x \ = \ 0$$

$$\color{red}{5 \ log \ 12 \ - \ 25x \ = \ 0 \ ⇒ \ 5 \ log \ 12 \ = \ 25x \ ⇒ \ x \ = \ \frac{5 \ log \ 12}{25} \ = \ \frac{log \ 12}{5}}$$

8) Solve: $$log_7 \ (-2x \ + \ 3) \ - \ 7 \ = \ 5$$

$$\color{red}{log_7 \ (-2x \ + \ 3) \ - \ 7 \ = \ 5 \ ⇒ \ log_7 \ (-2x \ + \ 3) \ = \ 2 \ ⇒ \ -2x \ + \ 3 \ = \ 7^2 \ ⇒ }$$ $$\color{red}{ \ -2x \ = \ 46 \ ⇒}$$ $$\color{red}{x \ = \ -23}$$

9) Solve: $$log_5 \ (4x \ + \ 5) \ = \ 3$$

$$\color{red}{log_5 \ (4x \ + \ 5) \ = \ 3 \ ⇒ \ (4x \ + \ 5) \ = \ 5^3 \ ⇒ \ 4x \ = \ 120 \ ⇒ \ x \ = \ 30}$$

10) Solve: $$log_9 \ (2x \ + \ 1) \ + \ log_9 \ 5x \ = \ 2$$

$$\color{red}{log_9 \ (2x \ + \ 1) \ + \ log_9 \ 5x \ = \ 2 \ ⇒ \ log_9 \ (2x \ + \ 1) \times 5x \ = \ 2 \ ⇒ \ 10x^2 \ + \ 5x \ = \ 9^2 \ ⇒}$$ $$\color{red}{x_1 \ = \ \frac{\sqrt{3265}}{20} \ - \ \frac{1}{4} \ ≈ \ 2.607 \ , \ x_2 \ = \ -\frac{\sqrt{3265}}{20} \ - \ \frac{1}{4} \ ≈ \ 3.107}$$

## Solving Logarithmic Equations Practice Quiz

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