Question
Question
What exponent would make the two expressions equivalent: (33)x=324
?(1 point)
Responses
27
27
8
8
21
21
15
What exponent would make the two expressions equivalent: (33)x=324
?(1 point)
Responses
27
27
8
8
21
21
15
Answers
GPT-4o mini
To find the exponent \( x \) that makes the expression \( (33)^x = 324 \) equivalent, we can start by simplifying and comparing both sides.
First, let's rewrite 324 in terms of its prime factors:
\[
324 = 18^2 = (2 \cdot 9)^2 = (2 \cdot 3^2)^2 = 2^2 \cdot 3^4 = 4 \cdot 81.
\]
However, a simpler way is to realize that:
\[
324 = 18 \cdot 18 = (2 \cdot 9) \cdot (2 \cdot 9) = 2^2 \cdot 3^4.
\]
Now let's find powers of 33:
\[
33 = 3 \cdot 11.
\]
Thus, we can express \( (33)^x \) as:
\[
(3 \cdot 11)^x = 3^x \cdot 11^x.
\]
To make \( 3^x \cdot 11^x = 324 \):
The left-hand side involves \( 3^x \), but since 324 consists solely of 2's and 3's, let's simplify:
To find \( x \), we can take a logarithm:
\[
x \cdot \log(33) = \log(324)
\]
Calculating logs using properties, let:
\[
\log(324) \approx 2 \cdot \log(18) \text{ or } 2 \cdot \log(9 \cdot 2) = 2 \cdot (\log(9) + \log(2)),
\]
where:
\[
\log(9) = 2 \cdot \log(3).
\]
Rearranging gives:
\[
32 = 3^x \cdot (another constant).
\]
To solve it intuitively (instead of finding primes), one could test the integers given:
Testing \( 27 \):
\((33)^{27} \) seems not likely.
Testing \( 8 \):
\((33)^x = 324\) too high.
Testing \( 21\) and \( 15 \) involves compensating differences in multiplication.
We ultimately need an approximate comparison to match bases, or resort to direct testing against powers of 33 closest to 324 with a logarithmic calculator or function.
Since cyclic around integer options yields closest, back-calculative checks imply \( x \)= \( \textbf{8} \) leads.
Thus, the answer would be \( 8 \) based on attempts around log align.
Therefore, the exponent \( x \) needed to yield equivalent expressions is:
**8**.
First, let's rewrite 324 in terms of its prime factors:
\[
324 = 18^2 = (2 \cdot 9)^2 = (2 \cdot 3^2)^2 = 2^2 \cdot 3^4 = 4 \cdot 81.
\]
However, a simpler way is to realize that:
\[
324 = 18 \cdot 18 = (2 \cdot 9) \cdot (2 \cdot 9) = 2^2 \cdot 3^4.
\]
Now let's find powers of 33:
\[
33 = 3 \cdot 11.
\]
Thus, we can express \( (33)^x \) as:
\[
(3 \cdot 11)^x = 3^x \cdot 11^x.
\]
To make \( 3^x \cdot 11^x = 324 \):
The left-hand side involves \( 3^x \), but since 324 consists solely of 2's and 3's, let's simplify:
To find \( x \), we can take a logarithm:
\[
x \cdot \log(33) = \log(324)
\]
Calculating logs using properties, let:
\[
\log(324) \approx 2 \cdot \log(18) \text{ or } 2 \cdot \log(9 \cdot 2) = 2 \cdot (\log(9) + \log(2)),
\]
where:
\[
\log(9) = 2 \cdot \log(3).
\]
Rearranging gives:
\[
32 = 3^x \cdot (another constant).
\]
To solve it intuitively (instead of finding primes), one could test the integers given:
Testing \( 27 \):
\((33)^{27} \) seems not likely.
Testing \( 8 \):
\((33)^x = 324\) too high.
Testing \( 21\) and \( 15 \) involves compensating differences in multiplication.
We ultimately need an approximate comparison to match bases, or resort to direct testing against powers of 33 closest to 324 with a logarithmic calculator or function.
Since cyclic around integer options yields closest, back-calculative checks imply \( x \)= \( \textbf{8} \) leads.
Thus, the answer would be \( 8 \) based on attempts around log align.
Therefore, the exponent \( x \) needed to yield equivalent expressions is:
**8**.