Find the domain of the following functions.
`f(x)=sin^(-1)log_(2)""(x)/(3)`

Answer

To find the domain of the function \( f(x) = \sin^{-1} \left( \frac{\log_2 x}{3} \right) \), we need to ensure that the input to the sine inverse function is within its valid range. The sine inverse function, \( \sin^{-1}(y) \), is defined for values of \( y \) such that:

\[ -1 \leq y \leq 1 \]

In our case, we have:

\[ y = \frac{\log_2 x}{3} \]

Thus, we need to solve the following inequalities:

\[ -1 \leq \frac{\log_2 x}{3} \leq 1 \]

### Step 1: Solve the left inequality

Starting with the left part of the inequality:

\[ -1 \leq \frac{\log_2 x}{3} \]

Multiplying both sides by 3 (since 3 is positive, the inequality sign remains the same):

\[ -3 \leq \log_2 x \]

Now, converting the logarithmic inequality to exponential form:

\[ x \geq 2^{-3} \]

Calculating \( 2^{-3} \):

\[ x \geq \frac{1}{8} \]

### Step 2: Solve the right inequality

Now, we solve the right part of the inequality:

\[ \frac{\log_2 x}{3} \leq 1 \]

Multiplying both sides by 3:

\[ \log_2 x \leq 3 \]

Converting this logarithmic inequality to exponential form:

\[ x \leq 2^3 \]

Calculating \( 2^3 \):

\[ x \leq 8 \]

### Step 3: Combine the results

Now we have two inequalities:

1. \( x \geq \frac{1}{8} \)
2. \( x \leq 8 \)

Combining these gives us the domain:

\[ \frac{1}{8} \leq x \leq 8 \]

### Final Result

Thus, the domain of the function \( f(x) \) is:

\[ x \in \left[ \frac{1}{8}, 8 \right] \]