To solve the equation \( (33)^x = 324 \), we can take the logarithm of both sides.
Let's express it in logarithmic form:
\[ x = \frac{\log(324)}{\log(33)} \]
Now we can calculate \( \log(324) \) and \( \log(33) \):
- First, we recognize that \( 324 = 18^2 = (2 \cdot 3^2)^2 = 2^2 \cdot 3^4 \).
You can approximate the logarithm values or compute them:
- \( \log(324) \approx 2.5104 \)
- \( \log(33) \approx 1.5185 \)
Calculating:
\[ x \approx \frac{2.5104}{1.5185} \approx 1.65 \]
Since none of the provided options (21, 8, 27, 15) correctly represent this exponent, I suspect you might have misinterpreted the options or the equation form for equivalence.
To check specific practical integer or rational values, we can also raise \( 33 \) to these powers and look for equivalence with \( 324 \):
- \( (33)^2 = 1089 \) (too large)
- \( (33)^1 = 33 \) (too small)
- \( (33)^8 \) or \( (33)^15 \) or \( (33)^{27} \) will greatly exceed \( 324 \).
This means we should review your options for correctness or simplicity. The value \( x \) is indeed between \( 1 \) and \( 2 \), indicating no integer from the selections truly fits unless there’s additional context or another logarithmic approach desired.
Thus the previous logic holds - \( x \) near \( 1.65 \) is non-existent in your choice scope, leading to a conclusion that none of the mention points (21, 8, 27, 15) could make this statement true.