Prove that $x=1$ for $x+3^x = 4.$ How can I do this equation with exponentials? 
my intent is:
$\ln(3^x) = \ln(4-x)$, then $x\ln(3) = \ln(4-x)$... $x=1$?
Help, please.
 A: You can check by inspection that $x=1$ is a solution.
To check there are no more solutions, consider $4 - x = 3^x$. For $x < 1$, $3^x < 3$ while $4-x > 3$, so there can be no solution. Similarly for $x > 1$, $3^x > 3$ and $4-x < 3$, so no solution.
A: As @Deepak points out, unless you are willing to introduce a particular special function known as the Lambert W function it is not possible to find a closed-form solution to your equation in terms of familiar elementary functions.
To solve your equation in terms of the Lambert W function, $\text{W} (x)$, one first needs to write it in the form of its defining equation given by
$$\text{W} (x) e^{\text{W} (x)} = x.$$
Rearranging your equation we have
\begin{align*}
x + 3^x &= 4\\
e^{x \ln 3} &= 4 - x\\
(4 - x) e^{-x \ln 3} &= 1\\
(4 \ln 3 - x \ln 3) e^{-x \ln 3} &= \ln 3\\
(4 \ln 3 - x \ln 3) e^{4 \ln 3 - x \ln 3} &= 3^4 \cdot \ln 3.
\end{align*}
As the above equation is now in the form for the defining equation for the Lambert W function, in terms of this function its solution is given by
$$4 \ln 3 - x \ln 3 = \text{W}_\nu (3^4 \cdot \ln 3),$$
or
$$x = \frac{4 \ln 3 - \text{W}_\nu (3^4 \cdot \ln 3)}{\ln 3}.$$
Here $\nu$ denotes the branch of the Lambert W function. For real solutions, only the principal branch of $\nu = 0$ and the secondary real branch of $\nu = -1$ can be selected. However, as the argument for the Lambert W function is positive only the principal branch is chosen. Thus there is only one real solution to the equation and it is given by
$$x = \frac{4 \ln 3 - \text{W}_0 (3^4 \cdot \ln 3)}{\ln 3}.$$
The above solution can be simplified using the following simplification rule for the Lambert W function
$$\text{W}_0 (x \ln x) = \ln x, \quad x \geqslant \frac{1}{e}.$$
Observing that
$$\text{W}_0 (3^4 \cdot \ln 3) = \text{W}_0 (3^3 \cdot \ln (3^3)) = \ln (3^3) = 3 \ln 3,$$
the final solution reduces to
$$x = \frac{4 \ln (3) - 3 \ln (3)}{\ln 3} = 1,$$
as expected. 
