Which of the following options represents the value of `logsqrt(128)` to the base `0.625` ?

Options

A. `(2+log_(8)2)/(log_(8)5-1)`
B. `(log_(8)128)/(2log_(8)0.625)`
C. `(2+log_(8)2)/(2(log_(8)5-1))`
D. Both (b) and (c )

Answer

To solve the problem of finding the value of \( \log_{0.625}(\sqrt{128}) \), we can follow these steps:

### Step 1: Rewrite the logarithm
We start with the expression:
\[ \log_{0.625}(\sqrt{128}) \]
Using the property of logarithms that states \( \log_b(a^c) = c \cdot \log_b(a) \), we can rewrite the square root as an exponent:
\[ \sqrt{128} = 128^{1/2} \]
Thus, we can rewrite the logarithm as:
\[ \log_{0.625}(\sqrt{128}) = \log_{0.625}(128^{1/2}) = \frac{1}{2} \log_{0.625}(128) \]

### Step 2: Change of base formula
Next, we use the change of base formula for logarithms, which states that:
\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \]
for any base \( k \). We will use base 10 (or natural logarithm) for convenience:
\[ \log_{0.625}(128) = \frac{\log(128)}{\log(0.625)} \]
Substituting this back into our expression gives:
\[ \log_{0.625}(\sqrt{128}) = \frac{1}{2} \cdot \frac{\log(128)}{\log(0.625)} \]

### Step 3: Calculate the logarithms
Now we need to find the values of \( \log(128) \) and \( \log(0.625) \).
- We know that \( 128 = 2^7 \), so:
\[ \log(128) = \log(2^7) = 7 \cdot \log(2) \]
- For \( 0.625 \), we can express it as \( 0.625 = \frac{5}{8} = \frac{5}{2^3} \). Thus:
\[ \log(0.625) = \log(5) - \log(2^3) = \log(5) - 3 \cdot \log(2) \]

### Step 4: Substitute back
Now we substitute these values back into our expression:
\[ \log_{0.625}(\sqrt{128}) = \frac{1}{2} \cdot \frac{7 \cdot \log(2)}{\log(5) - 3 \cdot \log(2)} \]

### Step 5: Simplify the expression
This gives us the final expression for \( \log_{0.625}(\sqrt{128}) \):
\[ \log_{0.625}(\sqrt{128}) = \frac{7 \cdot \log(2)}{2(\log(5) - 3 \cdot \log(2))} \]

### Conclusion
Now we can evaluate this expression using the values of \( \log(2) \) and \( \log(5) \) if needed, or we can leave it in this form for comparison with the options provided.