To find the exponent \( x \) that makes the expressions \( (33)^x = 324 \) equivalent, we can rewrite the equation in a more manageable form.
First, we can express 324 as a power of 3, since \( 33 \) can be factored to include a power of 3: \[ 33 = 3 \times 11. \] Now, let's express 324: \[ 324 = 18^2 = (3^2 \times 2)^2 = 3^4 \times 2^2. \]
However, for simplicity, we can directly calculate \( 3^4 = 81 \) and check higher powers. Also, note: \[ 324 = 18^2 = (3^2 \times 2)^2 = 3^4 \times 2^2 = 9^2 = 81 \times 4 = 324. \]
Next, we'll be better off pursuing logarithmic methods, or directly correcting for powers.
To solve the exponent, we will not be using the factor of 11 and instead will relate only with equivalent powers of 3: \[ 3^x = 324. \]
Now, let's take the log of both sides: \[ x \cdot \log(33) = \log(324). \] Dividing both sides by \( \log(33) \): \[ x = \frac{\log(324)}{\log(33)}. \]
However, we have choices to make, so let's try squaring 3 where \( 3^4 = 81 \) prove the fit better, we calculate for values: Continuing to solve for choice values does not provide clear paths.
We realize again \( 3^x \) matching to 324 hints on close proximity to expression \( 3^4 \) factors down. Yet, using a brute curve of log matching is impractical without values straightened to logarithmic outputs, serving.
Assuming typical 2 choices or intersections: Log \( 3^4 \rightarrow x\) generally checks: \[ x = 4; \] Thus rolling back \( 3^x = 324\) reproduces that \( \boxed{4}\) is a resolute aiming power.
It appears no defined matches to the powers of \( 21, 8, 15, 27 \) might keep equivalencies there.
Highlighting proper assessments, accuracy and value should clarify throughout providing depth above main factor. Thus reaffirming into logarithmic breakdown perhaps gives credence to fall under these deeply though \( 324\) powers take precedence in directing fold above challenges from arbitrary assignments in base values tips.