The domain of sin-1 (log (x/3)] is. .

The correct option is [1, 9].

To find the domain of the function given by:
$$f(x)={\sin }^{-1}\left[\log \left(\frac{x}{3}\right)\right]$$
we must identify all values of $x$ for which the function is mathematically defined. We do this by analyzing the constraints imposed by each part of the expression step-by-step.

Step 1: Constraint of the arcsine function
The domain of the standard inverse sine function, ${\sin }^{-1}(\theta )$, is restricted to the closed interval $[-1,1]$. This means the argument inside the arcsine must satisfy the inequality:
$$-1\le \log \left(\frac{x}{3}\right)\le 1$$
Note: In standard mathematical contexts where the base of the logarithm is not explicitly written, it is assumed to be either natural (base $e$) or common (base 10) depending on convention. For the interval to match $[1,9]$, the base of the logarithm must be 3. Let us denote the function as ${\log }_3(x/3)$.
Thus, our inequality becomes:
$$-1\le {\log }_3\left(\frac{x}{3}\right)\le 1$$

Step 2: Solving the inequality for $x$
Since the logarithmic base is 3 (which is greater than 1), the logarithmic function is strictly increasing. We can eliminate the logarithm by applying the base 3 exponential to all parts of the inequality without reversing the inequality signs:
$$3^{-1}\le \frac{x}{3}\le 3^1$$
Simplifying the powers of 3:
$$\frac{1}{3}\le \frac{x}{3}\le 3$$

Step 3: Finding the range of $x$
To isolate $x$, we multiply the entire inequality by 3 (which is positive, so the inequality directions remain unchanged):
$$1\le x\le 9$$
Since $x$ lies in the range $[1,9]$, the argument of the logarithm $(x/3)$ will lie in $[1/3,3]$, which is strictly positive, satisfying the natural domain constraint of the logarithm itself ($x>0$).

Conclusion
Combining the constraints, the domain of the function is the set of all real numbers from 1 to 9, inclusive.
Therefore, the domain is written in interval notation as [1, 9].