# How do you solve the equation log(x)+log(x-8)=log9?

Feb 9, 2018

$x = 9$

#### Explanation:

Use ${\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n m\right)$.

$\log \left(x \left(x - 8\right)\right) = \log 9$

Therefore:

${x}^{2} - 8 x = 9$

${x}^{2} - 8 x - 9 = 0$

$\left(x - 9\right) \left(x + 1\right) = 0$

$x = 9 \mathmr{and} - 1$

But $x = - 1$ is impossible because the logarithm function is undefined. If you test $x = 9$ in the equation you will immediately see it works as $\log \left(9\right) + \log \left(9 - 8\right) = \log \left(9\right) + \log \left(1\right) = \log \left(9\right) + 0 = \log \left(9\right)$

Hopefully this helps!