# How do you solve 8^y = 4^(2x+3) and log2y=log2x+4?

The first equation becomes

${8}^{y} = {4}^{2 x + 3} \implies {\left({2}^{3}\right)}^{y} = {2}^{2 \cdot \left(2 x + 3\right)} \implies 3 \cdot y = 4 x + 6$

The second equation becomes

log2y=log2x+4=>log((2y)/(2x))=4=>log(y/x)=4=> y/x=10^4=>y=10^4*x

Now the system of equations become

$3 y = 4 x + 6$
$y = {10}^{4} \cdot x$

which has solutions

$x = \frac{3}{14998}$, $y = \frac{15000}{7499}$