# How do you solve 8^(x-y)=3^x and 8^y=2^(x-3)?

$x = \log \frac{8}{\log \left(\frac{3}{4}\right)} , y = \frac{\log \frac{8}{\log \left(\frac{3}{4}\right)} - 3}{3}$
From ${8}^{y} = {2}^{x - 3}$ knowing that $8 = {2}^{3}$ we can write
${2}^{3 y} = {2}^{x - 3}$ so we conclude that $3 y = x - 3$. Operating now on ${8}^{x - y} = {3}^{x} \to {2}^{3 \left(x - y\right)} = {3}^{x}$ but $3 y = x - 3$ substituting we get
${2}^{2 x + 3} = {3}^{x} \to 8 \times {4}^{x} = {3}^{x} \to {\left(\frac{3}{4}\right)}^{x} = 8$. Finally $x = \log \frac{8}{\log \left(\frac{3}{4}\right)}$. Solving for $y$ we get $y = \frac{\log \frac{8}{\log \left(\frac{3}{4}\right)} - 3}{3}$