Which term is larger, $$ \log _{25}\left(\frac{1}{625}\right) $$ or $$ \log _{16}\left(\frac{1}{64}\right) $$ ? (1 point) Neither logarithm is defined for values less than 1 . $$ \log _{25}\left(\frac{1}{625}\right)=\log _{16}\left(\frac{1}{64}\right) $$ $$ \log _{25}\left(\frac{1}{625}\right)\log _{16}\left(\frac{1}{64}\right) $$ Check answer $$ ) $$ Remaining Attempts : 3

Question

Which term is larger, $$ \log _{25}\left(\frac{1}{625}\right) $$ or $$ \log _{16}\left(\frac{1}{64}\right) $$ ? (1 point) Neither logarithm is defined for values less than 1 . $$ \log _{25}\left(\frac{1}{625}\right)=\log _{16}\left(\frac{1}{64}\right) $$ $$ \log _{25}\left(\frac{1}{625}\right)>\log _{16}\left(\frac{1}{64}\right) $$ Check answer $$ ) $$ Remaining Attempts : 3

UpStudy AI · Accepted Answer

Let's evaluate both logarithms:

1. **$$ \log_{25}\left(\frac{1}{625}\right) $$**
   - Since $$ 25 = 5^2 $$ and $$ 625 = 5^4 $$, we can rewrite:
   $$
   \log_{25}\left(\frac{1}{625}\right) = \log_{5^2}\left(5^{-4}\right) = \frac{-4}{2} = -2
   $$

2. **$$ \log_{16}\left(\frac{1}{64}\right) $$**
   - Since $$ 16 = 2^4 $$ and $$ 64 = 2^6 $$, we can rewrite:
   $$
   \log_{16}\left(\frac{1}{64}\right) = \log_{2^4}\left(2^{-6}\right) = \frac{-6}{4} = -1.5
   $$

Comparing the two results:
$$
-2 < -1.5
$$
This means:
$$
\log_{25}\left(\frac{1}{625}\right) < \log_{16}\left(\frac{1}{64}\right)
$$

However, based on the provided options, the closest correct statement is that $$ \log_{25}\left(\frac{1}{625}\right) $$ is **less than** $$ \log_{16}\left(\frac{1}{64}\right) $$.

**Answer:** $$ \log _{25}\left(\frac{1}{625}\right)<\log _{16}\left(\frac{1}{64}\right) $$
$$ \log_{25}\left(\frac{1}{625}\right) < \log_{16}\left(\frac{1}{64}\right) $$

Which term is larger, or ?
(1 point)
Neither logarithm is defined for values less than 1 .

Check answer Remaining Attempts : 3

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Which term is larger, or ? (1 point) Neither logarithm is defined for values less than 1 . Check answer Remaining Attempts : 3