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Answer :
To solve the problem of finding the equivalent expression to [tex]\(\frac{\log_9 128}{\log_2 16}\)[/tex], let's break it down step-by-step using properties of logarithms.
1. Change of Base Formula:
The change of base formula states that [tex]\(\log_b a = \frac{\log_c a}{\log_c b}\)[/tex], where [tex]\(c\)[/tex] is any positive number. We'll use this to express [tex]\(\log_9 128\)[/tex] in terms of base 2.
2. Simplify [tex]\(\log_9 128\)[/tex]:
Using the change of base formula, [tex]\(\log_9 128 = \frac{\log_2 128}{\log_2 9}\)[/tex].
3. Simplify [tex]\(\log_2 16\)[/tex]:
We know that [tex]\(16 = 2^4\)[/tex], so [tex]\(\log_2 16 = 4\)[/tex].
4. Combine Both Parts:
Now we substitute these into the original expression:
[tex]\[
\frac{\log_9 128}{\log_2 16} = \frac{\frac{\log_2 128}{\log_2 9}}{4} = \frac{\log_2 128}{4 \cdot \log_2 9}
\][/tex]
5. Simplify Further:
Since the expression [tex]\(\frac{\log_2 128}{4 \cdot \log_2 9}\)[/tex] simplifies to [tex]\(\log_2 128\)[/tex] divided by 4, we can further simplify it using the power rule for logarithms:
[tex]\[
\frac{1}{4} \cdot \log_2 128 = \log_2 \left(128^{1/4}\right)
\][/tex]
Since this is equivalent to [tex]\(\log_2 128\)[/tex], we recognize that the expression simplifies to [tex]\(\log_2 128\)[/tex].
Therefore, the expression [tex]\(\frac{\log_9 128}{\log_2 16}\)[/tex] is equivalent to [tex]\(\log_2 128\)[/tex]. This corresponds to the first option in the choices provided.
1. Change of Base Formula:
The change of base formula states that [tex]\(\log_b a = \frac{\log_c a}{\log_c b}\)[/tex], where [tex]\(c\)[/tex] is any positive number. We'll use this to express [tex]\(\log_9 128\)[/tex] in terms of base 2.
2. Simplify [tex]\(\log_9 128\)[/tex]:
Using the change of base formula, [tex]\(\log_9 128 = \frac{\log_2 128}{\log_2 9}\)[/tex].
3. Simplify [tex]\(\log_2 16\)[/tex]:
We know that [tex]\(16 = 2^4\)[/tex], so [tex]\(\log_2 16 = 4\)[/tex].
4. Combine Both Parts:
Now we substitute these into the original expression:
[tex]\[
\frac{\log_9 128}{\log_2 16} = \frac{\frac{\log_2 128}{\log_2 9}}{4} = \frac{\log_2 128}{4 \cdot \log_2 9}
\][/tex]
5. Simplify Further:
Since the expression [tex]\(\frac{\log_2 128}{4 \cdot \log_2 9}\)[/tex] simplifies to [tex]\(\log_2 128\)[/tex] divided by 4, we can further simplify it using the power rule for logarithms:
[tex]\[
\frac{1}{4} \cdot \log_2 128 = \log_2 \left(128^{1/4}\right)
\][/tex]
Since this is equivalent to [tex]\(\log_2 128\)[/tex], we recognize that the expression simplifies to [tex]\(\log_2 128\)[/tex].
Therefore, the expression [tex]\(\frac{\log_9 128}{\log_2 16}\)[/tex] is equivalent to [tex]\(\log_2 128\)[/tex]. This corresponds to the first option in the choices provided.
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