To solve for $f^{-1}(16)$, we need to first understand the function $f(x) = 2^{\log_3(x)}$.

Step 1: Find $f(x)$ in a simpler form

We know: $f(x) = 2^{\log_3(x)}$

To understand this function, let's rewrite the logarithm using a change of base formula. The change of base formula states:

$\log_3(x) = \frac{\log_2(x)}{\log_2(3)}$

Substituting this into the expression for $f(x)$:

$f(x) = 2^{\log_3(x)} = 2^{\frac{\log_2(x)}{\log_2(3)}}$

Now, using the exponent rule $a^{b \cdot c} = (a^b)^c$, we can rewrite $f(x)$:

$f(x) = \left( 2^{\log_2(x)} \right)^{\frac{1}{\log_2(3)}}$

Since $2^{\log_2(x)} = x$, we get:

$f(x) = x^{\frac{1}{\log_2(3)}}$

Step 2: Find $f^{-1}(x)$

To find $f^{-1}(x)$, we express $x$ as a function of $y$:

$y = x^{\frac{1}{\log_2(3)}}$

Now, solving for $x$ gives:

$y = x^{\frac{1}{\log_2(3)}}$ $y^{\log_2(3)} = x$

So, $f^{-1}(y) = y^{\log_2(3)}$.

Step 3: Evaluate $f^{-1}(16)$

Now, we substitute $y = 16$ into $f^{-1}(y)$:

$f^{-1}(16) = 16^{\log_2(3)}$

Since $16 = 2^4$:

$f^{-1}(16) = (2^4)^{\log_2(3)}$ $= 2^{4 \cdot \log_2(3)}$ $= 2^{\log_2(3^4)}$ $= 3^4$ $= 81$

Therefore, the value of $f^{-1}(16)$ is $\boxed{81}$.

Answer:

(c) 81

Would you like more details, or do you have any questions?

Here are five related questions to expand on this topic:

1. What is the domain and range of the function $f(x) = 2^{\log_3(x)}$?
2. How would you find the inverse of a logarithmic function in general?
3. Can you express $f(x) = 2^{\log_3(x)}$ using natural logarithms?
4. What is the graphical interpretation of finding the inverse of a function?
5. How would you solve for $f^{-1}(32)$ if $f(x) = 2^{\log_3(x)}$?

Tip: Understanding the change of base formula for logarithms can simplify many calculations and is a valuable tool when dealing with functions involving different bases.