If \( 4^{x-1}=8 \), what is \( x \) ? \( \begin{array}{lllll}\text { A } 2 \frac{1}{4} & \text { B } 2 \frac{1}{2} & \text { C } 3 & \text { D) } 5 & \text { E. } 33\end{array} \)

Let's solve the equation \( 4^{x-1} = 8 \) by rewriting both sides in terms of powers of 2. First, notice that \( 4 = 2^2 \) and \( 8 = 2^3 \). This allows us to express the equation as: \[ (2^2)^{x-1} = 2^3 \] By applying the power of a power rule, we can simplify the left side: \[ 2^{2(x-1)} = 2^3 \] With the bases being the same, we can set the exponents equal to each other: \[ 2(x-1) = 3 \] Solving for \( x \) gives: \[ 2x - 2 = 3 \\ 2x = 5 \\ x = \frac{5}{2} = 2.5 \] Now, let's convert \( 2.5 \) into a mixed fraction: \[ x = 2 \frac{1}{2} \] Thus, the answer is \( \text{B) } 2 \frac{1}{2} \). To double-check our answer, we can substitute \( x = 2.5 \) back into the original equation: \[ 4^{2.5 - 1} = 4^{1.5} = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \] The calculation holds true, confirming that our solution is correct. So the answer is \( \text{B) } 2 \frac{1}{2} \).