Which is greater `log_(2)3` or `log_(1//2)5....`

Answer

To determine which is greater between \( \log_{2}3 \) and \( \log_{\frac{1}{2}}5 \), we can use properties of logarithms to simplify our comparison.

### Step-by-Step Solution:

1. **Identify the logarithmic expressions**:
We have two expressions to compare:
\[ A = \log_{2}3 \]
\[ B = \log_{\frac{1}{2}}5 \]

2. **Convert \( B \) to a base of 2**:
We can use the change of base formula for logarithms:
\[ \log_{a}b = \frac{\log_{c}b}{\log_{c}a} \]
Here, we will convert \( B \) to base 2:
\[ B = \log_{\frac{1}{2}}5 = \frac{\log_{2}5}{\log_{2}\left(\frac{1}{2}\right)} \]
Since \( \log_{2}\left(\frac{1}{2}\right) = -1 \), we have:
\[ B = \frac{\log_{2}5}{-1} = -\log_{2}5 \]

3. **Compare \( A \) and \( B \)**:
Now we can rewrite our comparison:
\[ A = \log_{2}3 \]
\[ B = -\log_{2}5 \]
Thus, we need to compare:
\[ \log_{2}3 \quad \text{and} \quad -\log_{2}5 \]

4. **Rearranging the comparison**:
We can rewrite the comparison as:
\[ \log_{2}3 + \log_{2}5 > 0 \]
This simplifies to:
\[ \log_{2}(3 \times 5) > 0 \]
\[ \log_{2}15 > 0 \]

5. **Conclusion**:
Since \( \log_{2}15 > 0 \), this implies that:
\[ \log_{2}3 > -\log_{2}5 \]
Therefore, we conclude that:
\[ \log_{2}3 > \log_{\frac{1}{2}}5 \]
Thus, the final answer is:
\[ \log_{2}3 \text{ is greater than } \log_{\frac{1}{2}}5 \]