# How do you solve 3 ^(5x) = 27^(2x + 1)?

Apr 14, 2016

Rearrange and take the logarithms to the base 3 of both sides, to yield $x = - 3$

#### Explanation:

Rewrite the right hand side:

${27}^{2 x + 1} = {3}^{3 \left(2 x + 1\right)} = {3}^{6 x + 3}$

Now the overall equation is:

${3}^{5 x} = {3}^{6 x + 3}$

Taking the log to the base 3 of both sides:

$5 x = 6 x + 3$

$x = - 3$