To solve the equation \( 32^{x-1} = 27 \), we start by rewriting both sides using exponential forms.
First, express 32 and 27 as powers of 2 and 3:
- \( 32 = 2^5 \)
- \( 27 = 3^3 \)
So, we rewrite the equation: \[ (2^5)^{x-1} = 3^3 \] This simplifies to: \[ 2^{5(x-1)} = 3^3 \] Next, take the logarithm on both sides: \[ \log(2^{5(x-1)}) = \log(3^3) \] Using the logarithmic identity \( \log(a^b) = b\log(a) \): \[ 5(x-1) \log(2) = 3 \log(3) \] Now, solve for \( x \): \[ 5(x-1) = \frac{3 \log(3)}{\log(2)} \] \[ x-1 = \frac{3 \log(3)}{5 \log(2)} \] \[ x = 1 + \frac{3 \log(3)}{5 \log(2)} \]
To find an approximate value of \( x \), we can calculate:
- \( \log(2) \approx 0.3010 \)
- \( \log(3) \approx 0.4771 \)
Now calculate: \[ x \approx 1 + \frac{3 \times 0.4771}{5 \times 0.3010} \approx 1 + \frac{1.4313}{1.505} \approx 1 + 0.9503 \approx 1.9503 \]
The closest answer you provided from the options is 2. Therefore, the solution is approximately: \[ \boxed{2} \]