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

Jun 11, 2016

${\left({2}^{3}\right)}^{y} = {2}^{x - 3}$

${2}^{3 y} = {2}^{x - 3}$

We're now in equal bases, so we can eliminate the bases.

$3 y = x - 3$

$3 y + 3 = x$

${8}^{\left(3 y + 3\right) - y} = {3}^{3 y + 3}$

${8}^{2 y + 3} = {3}^{3 y + 3}$

This can be solved using the property ${n}^{m} = {a}^{b} \to \log {n}^{m} = \log {a}^{b}$

$\log {8}^{2 y + 3} = \log {3}^{3 y + 3}$

Now, we need to simplify using the property $\log {a}^{n} = n \log a$.

$\left(2 y + 3\right) \log 8 = \left(3 y + 3\right) \log 3$

$2 y \log 8 + 3 \log 8 = 3 y \log 3 + 3 \log 3$

$2 y \log 8 - 3 y \log 3 = 3 \log 3 - 3 \log 8$

$y \left(2 \log 8 - 3 \log 3\right) = 3 \log 3 - 3 \log 8$

Now, simplifying using the rule ${\log}_{a} \left(n\right) - {\log}_{a} \left(m\right) = {\log}_{a} \left(\frac{n}{m}\right)$:

$y = \frac{\log \left(\frac{27}{512}\right)}{\log \left(\frac{64}{27}\right)}$

Now, substituting back into the original equation to find x:

${2}^{3 y} = {2}^{x - 3}$

${2}^{3 \left(\frac{\log \left(\frac{27}{512}\right)}{\log \left(\frac{64}{27}\right)}\right)} = {2}^{x - 3}$

$3 \left(\frac{\log \left(\frac{27}{512}\right)}{\log \left(\frac{64}{27}\right)}\right) = x - 3$

$\frac{\log \left(\frac{19683}{134217728}\right)}{\log \left(\frac{64}{27}\right)} + 3 = x$

The solution set is therefore {(log(19683/134217728))/(log(64/27)) + 3, (log(27/512))/(log(64/27))} or {-7.23; -3.41}.

Hopefully this helps!