# Solve logarithmic equations

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.

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

logb X = Y

To solve it, you just have to write it in exponential form as X = bY

Example #1

log (2x + 1) = 3

log10 (2x + 1) = 3

After writing it in exponential form we get

2x + 1 = 103

2x + 1 = 1000

2x = 999

x = 499.5

Check x = 499.5

log10 (2 x 499.5 + 1)  = log10 (1000) = 3 since 103 = 1000

Example #2

log5 (30x - 10) - 2 = log5 (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.

log5 (30x - 10) - log5 (x + 6) - 2 = log5 (x + 6) - log5 (x + 6)

log5 (30x - 10) - log5 (x + 6) -2 = 0

log5 (30x - 10) - log5 (x + 6) -2 + 2 = 0 + 2

log5 (30x - 10) - log5 (x + 6)  =  2

log5 [ (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 = 52

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

log5 (30x - 10) - 2 = log5 (x + 6)

log5 (30 x 32 - 10) - 2 = log5 (32 + 6)

log5 (960 - 10) - 2 = log5 (32 + 6)

log5 (950) - 2 = log5 (38)

4.26 - 2 = 2.26

2.26 = 2.26

x = 32 is a good answer.

The second type you will learn to solve has the following format

logb X = logb Y

To solve it, you just have to see that logb X = logb Y if and only if X = Y

## More examples of how to solve logarithmic equations using logarithmic properties.

Example #3

log6 (2x - 4) + log6 (4) = log6 40

log6 4(2x - 4)  = log6 40

log6 8x - 16  = log6 40

8x - 16 = 40

8x = 40 + 16

8x = 56

x = 7

Check x = 7

log6 (2 x 7 - 4) + log6 (4) = log6 40

log6 (14 - 4) + log6 (4) = log6 40

log6 (10) + log6 (4) = log6 40

log6 (10 x 4) = log6 40

log6 (40) = log6 40

x is 7 is a good answer.

Example #4

log3 x + log3 (x + 3)  = log3 (2x + 6)

log3 x(x + 3)  = log3 (2x + 6)

log3 (x2 + 3x)  = log3 (2x + 6)

x2 + 3x  = 2x + 6

x2 + 3x - 2x - 6 = 0

x2 + x - 6 = 0

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

x = 2 and x  = -3

Check x = 2

log3 2 + log3 (2 + 3)  = log3 (2 x 2 + 6)

log3 2 + log3 (5)  = log3 (4 + 6)

log3 2(5)  = log3 (4 + 6)

log3 (10)  = log3 (10)

x  = 2  is a good answer.

Check x = -3

log3 -3 + log3 (-3 + 3)  = log3 (2 x -3 + 6)

log3 -3 + log3 0 = log3 0

Since
log3 -3 does not exist, x  = -3 cannot be answer.

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