To simplify the expression \( \log_2 128 \log_2 16 \), we can first calculate each logarithm:
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Calculate \( \log_2 128 \):
- \( 128 = 2^7 \), so \( \log_2 128 = 7 \).
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Calculate \( \log_2 16 \):
- \( 16 = 2^4 \), so \( \log_2 16 = 4 \).
Now we can substitute these values into the expression:
\[ \log_2 128 \cdot \log_2 16 = 7 \cdot 4 = 28 \]
Next, we need to find the expression equivalent to \( \log_2 128 \log_2 16 \).
We notice that:
\[ \log_2 128 = 7 \quad \Rightarrow \quad 2^7 = 128 \] \[ \log_2 16 = 4 \quad \Rightarrow \quad 2^4 = 16 \]
The term \( \log_2 128 \log_2 16 \) can also be expressed in terms of a single logarithm:
\[ \log_2 128 = \log_2 (16^4) = 4 \log_2 16 \]
Thus:
\[ \log_2 128 \log_2 16 = \log_2 (16^4) \cdot \log_2 16 \]
To express \( 7 \) as \( \log_4 128 \):
Using change of base:
\[ \log_2 128 = \frac{\log_4 128}{\log_4 2} \]
Since \( \log_4 2 = 1/2 \), we have: \[ \log_2 128 = 2 \log_4 128 \]
Thus, the corresponding expression matches with \( \log_4 128 \).
Given these calculations, the expression equivalent to \( \log_2 128 \log_2 16 \) from your options is:
log4128 logarithm subscript 4 baseline 128