If `bgt1, sintgt0,costgt0andlog_b(sint)=x," then "log_b(cost)` is equal to

Options

A. `1/2log_b(1-b^(2x))`
B. `2log(1-b^(x//2))`
C. `log_bsqrt(1-b^(2x))`
D. `sqrt(1-x^2)`

Explanation

To solve the problem, we need to find the value of \( \log_b(\cos t) \) given that \( \log_b(\sin t) = x \), where \( b > 1 \), \( \sin t > 0 \), and \( \cos t > 0 \).

### Step-by-step Solution:

1. **Start with the given information:**
\[ \log_b(\sin t) = x \]
This implies:
\[ \sin t = b^x \]

2. **Use the Pythagorean identity:**
We know from trigonometric identities that:
\[ \sin^2 t + \cos^2 t = 1 \]
Therefore, we can express \( \cos^2 t \) in terms of \( \sin t \):
\[ \cos^2 t = 1 - \sin^2 t \]

3. **Substitute \( \sin t \) into the identity:**
Substitute \( \sin t = b^x \):
\[ \cos^2 t = 1 - (b^x)^2 = 1 - b^{2x} \]

4. **Take the square root to find \( \cos t \):**
Since \( \cos t > 0 \), we take the positive root:
\[ \cos t = \sqrt{1 - b^{2x}} \]

5. **Express \( \log_b(\cos t) \):**
Now, we need to find \( \log_b(\cos t) \):
\[ \log_b(\cos t) = \log_b(\sqrt{1 - b^{2x}}) \]

6. **Use the logarithmic property:**
Using the property of logarithms that states \( \log_b(a^c) = c \cdot \log_b(a) \):
\[ \log_b(\cos t) = \frac{1}{2} \log_b(1 - b^{2x}) \]

7. **Final expression:**
Thus, we can express \( \log_b(\cos t) \) as:
\[ \log_b(\cos t) = \frac{1}{2} \log_b(1 - b^{2x}) \]

### Conclusion:
The final answer is:
\[ \log_b(\cos t) = \frac{1}{2} \log_b(1 - b^{2x}) \]