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

Apr 15, 2018

$x = 2$

#### Explanation:

Notice that all the logs are of the same base!

First of all ${\log}_{b} \left(x - 1\right) + {\log}_{b} \left(x + 2\right) = {\log}_{b} \left(\left(x - 1\right) \left(x + 2\right)\right)$

Therefore

${\log}_{b} \left(\left(x - 1\right) \left(x + 2\right)\right) = {\log}_{b} \left(8 - 2 x\right)$

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

$\left(x - 1\right) \left(x + 2\right) = 8 - 2 x$

This is a simple quadratic equation.

${x}^{2} + x - 2 = 8 - 2 x$
${x}^{2} + 3 x - 10 = 0$

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

${x}^{2} + 3 x = 10$

${x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} = 10 + {\left(\frac{3}{2}\right)}^{2}$

${\left(x + \frac{3}{2}\right)}^{2} = 10 + {\left(\frac{3}{2}\right)}^{2}$

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

--

${x}_{1} = \frac{7 - 3}{2} = 2$

${x}_{2} = \frac{- 7 - 3}{2} = - 5$

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