This lesson will discuss two ways to solve logarithmic equations. The method you use depends on the type of logarithmic equation you are trying to solve. Keep reading and you will learn how to solve the one below in a heartbeat!

The **first type** you will learn to solve has the following format

log_{b} X = Y

To solve it, you just have to write it in exponential form as X = b^{Y}**Example #1**

log (2x + 1) = 3

log_{10} (2x + 1) = 3

After writing it in exponential form we get

2x + 1 = 10^{3}

2x + 1 = 1000

2x = 999

x = 499.5**Check x = 499.5**log

Sometimes when you solve logarithmic equations, you need to put all the logarithms on one side of the equation. Example #2 shows how to do this.**Example #2**

log_{5} (30x - 10) - 2 = log_{5} (x + 6)

You can solve this one the same way you solved example #1.

Start by putting all logarithms on one side of the equation.

log_{5} (30x - 10) - log_{5} (x + 6) - 2 = log_{5} (x + 6) - log_{5} (x + 6)

log_{5} (30x - 10) - log_{5} (x + 6) -2 = 0

log_{5} (30x - 10) - log_{5} (x + 6) -2 + 2 = 0 + 2

log_{5} (30x - 10) - log_{5} (x + 6) = 2

log_{5} [ (30x - 10) / (x + 6) ] = 2

Now that it has the format that we want, we can just write it in exponential form.

30x - 10 / x + 6 = 5^{2}

30x - 10 / x + 6 = 25 ( After multiplying both sides by x + 6 )

30x - 10 = 25 (x + 6)

30x - 10 = 25x + 150

30x - 25x = 150 + 10

5x = 160

x = 32

**Check x = 32**

log_{5} (30x - 10) - 2 = log_{5} (x + 6)

log_{5} (30 x 32 - 10) - 2 = log_{5} (32 + 6)

log_{5} (960 - 10) - 2 = log_{5} (32 + 6)

log_{5} (950) - 2 = log_{5} (38)

4.26 - 2 = 2.26

2.26 = 2.26

x = 32 is a good answer.

Now, you will learn how to **solve logarithmic equations** that have the following **format**

log_{b} X = log_{b} Y

To solve it, you just have to see that log_{b} X = log_{b} Y if and only if X = Y

**Example #3**

log_{6} (2x - 4) + log_{6} (4) = log_{6} 40

log_{6} 4(2x - 4) = log_{6} 40

log_{6} 8x - 16 = log_{6} 40

8x - 16 = 40

8x = 40 + 16

8x = 56

x = 7**Check x = 7**

log_{6} (2 x 7 - 4) + log_{6} (4) = log_{6} 40

log_{6} (14 - 4) + log_{6} (4) = log_{6} 40

log_{6} (10) + log_{6} (4) = log_{6} 40

log_{6} (10 x 4) = log_{6} 40

log_{6} (40) = log_{6} 40

x is 7 is a good answer.

**Example #4**

log_{3} x + log_{3} (x + 3) = log_{3} (2x + 6)

log_{3} x(x + 3) = log_{3} (2x + 6)

log_{3} (x^{2} + 3x) = log_{3} (2x + 6)

x^{2} + 3x = 2x + 6

x^{2} + 3x - 2x - 6 = 0

x^{2} + x - 6 = 0

(x - 2) (x + 3) = 0

x = 2 and x = -3

**Check x = 2**

log_{3} 2 + log_{3} (2 + 3) = log_{3} (2 x 2 + 6)

log_{3} 2 + log_{3} (5) = log_{3} (4 + 6)

log_{3} 2(5) = log_{3} (4 + 6)

log_{3} (10) = log_{3} (10)

x = 2 is a good answer.

**Check x = -3**

log_{3} -3 + log_{3} (-3 + 3) = log_{3} (2 x -3 + 6)

log_{3} -3 + log_{3} 0 = log_{3} 0

Since log_{3} -3 does not exist, x = -3 cannot be an answer. It is an extraneous solution.

**Example #5**

log (x + 2) + log (x - 1) = 1

Use the product rule to rewrite the left side of the equation

log [(x + 2)×(x - 1)] = 1

log [(x^{2} - x + 2x - 2 )] = 1

log [(x^{2} + x - 2 )] = 1

Notice that log_{10} 10 = 1 since 10^{1} = 10

log_{10} [(x^{2} + x - 2 )] = log_{10} 10

x^{2} + x - 2 = 10

x^{2} + x - 2 - 10 = 10 - 10

x^{2} + x - 12 = 0

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

x = -4 and x = 3

Replace x = -4 and x = 3 into the original logarithmic equation to find the solution(s)

**Check x = -4**

log (x + 2) + log (x - 1) = 1

log (-4 + 2) + log (-4 - 1) = 1

log (-2) + log (-5) = 1

log (-2) and log (-5) do not exist since the logarithm of a negative number does not exist.

Therefore, x = -4 is an extraneous solution.

**Check x = 3**

log (x + 2) + log (x - 1) = 1

log (3 + 2) + log (3 - 1) = 1

log (5) + log (2) = 1

log(5×2) = 1

log 10 = 1

log_{10} 10 = 1

Since log_{10} 10 = 1 is a true statement, x = 3 is a solution.