Find the unique function $f(x)$ satisfying the following conditions:
\[
\begin{array}{r}
f^{\prime}(x)=3^{x} \\
f(0)=4
\end{array}
\]
\[
f(x)=
\]
国固

Step 1 ：We are given the derivative of the function $f(x)$ as $f'(x) = 3^x$ and a condition $f(0) = 4$.

Step 2 ：We can find the function $f(x)$ by integrating the derivative $f'(x) = 3^x$. The integral of $3^x$ with respect to $x$ is $\frac{3^x}{\log(3)} + C$, where $C$ is the constant of integration.

Step 3 ：We can find the value of $C$ by using the condition $f(0) = 4$. Substituting $x = 0$ into the function $f(x) = \frac{3^x}{\log(3)} + C$, we get $f(0) = \frac{3^0}{\log(3)} + C = 1/\log(3) + C$.

Step 4 ：Setting this equal to 4, we get $1/\log(3) + C = 4$. Solving for $C$, we find $C = 4 - 1/\log(3) = \frac{-1 + \log(81)}{\log(3)}$.

Step 5 ：Substituting $C$ back into the function $f(x)$, we get $f(x) = \frac{3^x}{\log(3)} + \frac{-1 + \log(81)}{\log(3)}$.

Step 6 ：\(\boxed{f(x) = \frac{3^x}{\log(3)} + \frac{-1 + \log(81)}{\log(3)}}\) is the unique function satisfying the given conditions.