Solve `log_((sin x))2log_((sin^(2)x))a=-1` stating any condition on a' that may be required for the existence of the solution.

Answer

To solve the equation \( \log_{\sin x} (2 \log_{\sin^2 x} a) = -1 \), we will follow a systematic approach.

### Step 1: Rewrite the equation
We start with the given equation:
\[ \log_{\sin x} (2 \log_{\sin^2 x} a) = -1 \]
Using the property of logarithms, we can rewrite this as:
\[ 2 \log_{\sin^2 x} a = \sin x^{-1} \]

### Step 2: Change of base formula
Using the change of base formula, we have:
\[ \log_{\sin^2 x} a = \frac{\log_2 a}{\log_2 (\sin^2 x)} = \frac{\log_2 a}{2 \log_2 (\sin x)} \]
Substituting this into our equation gives:
\[ 2 \cdot \frac{\log_2 a}{2 \log_2 (\sin x)} = -1 \]
This simplifies to:
\[ \frac{\log_2 a}{\log_2 (\sin x)} = -1 \]

### Step 3: Cross-multiplying
Cross-multiplying gives us:
\[ \log_2 a = -\log_2 (\sin x) \]
This can be rewritten as:
\[ \log_2 a = \log_2 \left( \frac{1}{\sin x} \right) \]

### Step 4: Exponentiating both sides
Exponentiating both sides results in:
\[ a = \frac{1}{\sin x} \]

### Step 5: Conditions for existence
For the logarithm to be defined, we need:
1. \( \sin x > 0 \) (since the logarithm is only defined for positive values).
2. \( a > 0 \) (since logarithm is defined for positive \( a \)).

### Step 6: Conclusion
Thus, the solution to the equation is:
\[ a = \frac{1}{\sin x} \]
with the conditions:
- \( \sin x > 0 \)
- \( a > 0 \)