# How do you solve 1/2(log_7x+log_7 8)=log_7 16?

Nov 26, 2016

${\log}_{7} x + {\log}_{7} 8 = {\log}_{7} \frac{16}{\frac{1}{2}}$

${\log}_{7} x + {\log}_{7} 8 = 2 {\log}_{7} 16$

We now apply the rules ${\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n \times m\right)$ and $a \log n = \log {n}^{a}$.

${\log}_{7} \left(x \times 8\right) = {\log}_{7} {16}^{2}$

${\log}_{7} 8 x = {\log}_{7} 256$

If $\log a = \log b$, then $a = b$.

$8 x = 256$

$x = 32$

Hopefully this helps!