# How do you solve log_6x+log_6(x-9)=2?

Jul 24, 2016

Use the log property ${\log}_{a} \left(n\right) + {\log}_{a} \left(m\right) = {\log}_{a} \left(n \times m\right)$:

$\implies {\log}_{6} \left(x \left(x - 9\right)\right) = 2$

$\implies {\log}_{6} \left({x}^{2} - 9 x\right) = 2$

$\implies {x}^{2} - 9 x = {6}^{2}$

$\implies {x}^{2} - 9 x = 36$

$\implies {x}^{2} - 9 x - 36 = 0$

$\implies \left(x - 12\right) \left(x + 3\right) = 0$

$\implies x = 12 \mathmr{and} - 3$

Checking in the original equation, we find that only $x = 12$ works. Hence, the solution set is $\left\{12\right\}$.

Hopefully this helps!