# How do you solve log(16+2b)=log(b^2-4b)?

Aug 22, 2016

$\left\{- 2 , 8\right\}$ is the solution set.

#### Explanation:

Use the property $\log \left(a\right) = \log \left(b\right) \to a = b$

$16 + 2 b = {b}^{2} - 4 b$

$0 = {b}^{2} - 6 b - 16$

$0 = \left(b - 8\right) \left(b + 2\right)$

$b = 8 \mathmr{and} - 2$

Checking in the original equation, both solutions work, hence the solution set of $\left\{- 2 , 8\right\}$.

Hopefully this helps!