# What is the solution set of 4[9^(x-1)] = 108?

${9}^{x - 1} = \frac{108}{4} = 27$
${9}^{x - 1} = {3}^{3}$
But $9 = {3}^{2}$ so:
${3}^{2 \left(x - 1\right)} = {3}^{3}$
$2 \left(x - 1\right) = 3$
$2 x - 2 = 3$
$x = \frac{5}{2}$