Question

How do you solve `log_b(x - 1) + log_b(x + 2) = log_b(8 - 2x)`?

Answer 1

`x=2`

Explanation:

Notice that all the logs are of the same base!

First of all `log_b(x-1)+log_b(x+2)=log_b((x-1)(x+2))`

Therefore

`log_b((x-1)(x+2))=log_b(8-2x)`

We can remove the logs since both sides are logarithms of the same base.

`(x-1)(x+2)=8-2x`

This is a simple quadratic equation.

`x^2 +x -2=8-2x`
`x^2 +3x -10=0`

This can be solved by completing the square (or by using the quadratic formula)

`x^2 +3x =10`

`x^2 +3x +(3/2)^2= 10+(3/2)^2`

`(x +3/2)^2 = 10+(3/2)^2`

`x +3/2 = ± sqrt{10+(3/2)^2`

`x = ± sqrt{10+(3/2)^2} - 3/2`

`x = ± sqrt{10+9/4} - 3/2`

`x = ± sqrt{49/4} - 3/2`

`x = ± 7/2 - 3/2`

`x = (-3± 7)/2`

--

`x_1 = (7-3)/2 = 2`

`x_2 = (-7-3)/2 = -5`

However, since you can't take the logarithm of a negative number, x cannot be less than 0. Therefore `x=2` is the solution