Which is greater `log_(4)5` or `log_(1/16)(1//25).....`

Answer

To compare \( \log_{4}5 \) and \( \log_{\frac{1}{16}}\left(\frac{1}{25}\right) \), we can follow these steps:

### Step 1: Define the logarithms
Let:
- \( a = \log_{4}5 \)
- \( b = \log_{\frac{1}{16}}\left(\frac{1}{25}\right) \)

### Step 2: Simplify \( b \)
Using the property of logarithms that states \( \log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)} \), we can rewrite \( b \):
\[ b = \log_{\frac{1}{16}}\left(\frac{1}{25}\right) = \log_{16^{-1}}(25^{-1}) = \log_{16}(25)^{-1} \]
This can be simplified further:
\[ b = -\log_{16}(25) \]

### Step 3: Rewrite \( \log_{16}(25) \)
We know that \( 16 = 4^2 \), so:
\[ \log_{16}(25) = \log_{4^2}(25) = \frac{1}{2} \log_{4}(25) \]
Thus, we have:
\[ b = -\frac{1}{2} \log_{4}(25) \]

### Step 4: Express \( \log_{4}(25) \)
Since \( 25 = 5^2 \), we can write:
\[ \log_{4}(25) = \log_{4}(5^2) = 2 \log_{4}(5) \]
Substituting this back into our expression for \( b \):
\[ b = -\frac{1}{2} \cdot 2 \log_{4}(5) = -\log_{4}(5) \]

### Step 5: Compare \( a \) and \( b \)
Now we have:
- \( a = \log_{4}(5) \)
- \( b = -\log_{4}(5) \)

This means:
\[ b = -a \]

### Step 6: Determine which is greater
Since \( a = \log_{4}(5) \) is positive (because \( 5 > 4 \)), it follows that:
\[ a > b \]

Thus, we conclude:
\[ \log_{4}(5) > \log_{\frac{1}{16}}\left(\frac{1}{25}\right) \]

### Final Conclusion
Therefore, \( \log_{4}(5) \) is greater than \( \log_{\frac{1}{16}}\left(\frac{1}{25}\right) \).

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