To find the exponent \(x\) that makes the expression \( (33)^x = 324 \) true, we can rewrite both sides in order to solve for \(x\).
-
Notice that \(324\) can be expressed as powers of \(3\): \[ 324 = 18^2 = (3 \cdot 2)^2 = 9 \cdot 36 = 3^4 \cdot 2^2 \] This does not lead directly to any power of \(33\), so let's check the other part.
-
We can calculate \(33^x\) more directly. Let's start with taking logarithms: \[ x \cdot \log(33) = \log(324) \] This implies \[ x = \frac{\log(324)}{\log(33)} \]
-
However, we can also reason in another way. Notice that \(324\) is also \(18^2\), which does not help too much with \(33\). Preventing an error here, we can look for potential matches.
-
Testing the choices given (15, 27, 8, or 21), we can calculate \(33^x\) for each of these values until we find \(324\) or a close approximation, or a common logarithm.
- \(33^1 = 33\)
- \(33^2 = 1089\) (too high)
-
There should be clarity in recognizing that:
- To find the exponent, we need to check if \(x\) can yield \(324\).
After evaluating base cases and considering, \(x\) that approximates to \(2\) would continue to yield \(33^x\) way beyond \(324\).
Given all computations and possibilities checked, none of the provided numbers directly fits as exponents yielding equal values unless normalized.
The analysis indicates potential \(x\) couldn’t be directly solved from the given options possibly due to domain considerations.
Thus, none of the responses seem fitting \(33^x = 324\) accurately since upon exponential conversion comparisons \( x≈1.5454\).
For the integer options namely \( 15\), \(27\), \(8\), or \(21\), none can yield \(324\) reliably through checking. Thus, the logic assists better conventional \(x \approx 1.5454\) only.
6 answers