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 $f^{\prime }(x)=3^x$ and a condition $f(0)=4$.

Step 2 ：We can find the function $f(x)$ by integrating the derivative $f^{\prime }(x)=3^x$. 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 ：$\boxed{{\displaystyle f(x)=\frac{3^x}{\log (3)}+\frac{-1+\log (81)}{\log (3)}}}$ is the unique function satisfying the given conditions.