# How do you solve 81^x = 27^(x + 2)?

Jul 22, 2015

More simply, $x = 6$

(logarithms are unnecessary in this case).

#### Explanation:

$81 = {3}^{4}$ and $27 = {3}^{3}$, so this equation can also be written as

${3}^{4 x} = {3}^{3 \left(x + 2\right)} = {3}^{3 x + 6}$.

Since the bases are now the same, we can equate exponents to get

$4 x = 3 x + 6$ so that

$x = 6$

(technically this equating of exponents requires the fact that exponential functions (with base not equal to 1) are one-to-one functions).