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

1 Answer
Apr 15, 2018

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