Simple Logarithms Equation $$3^x = 3 - x$$

I have to prove that only one solution exists, and then find that one solution.

My approach has been the following:

$$\log 3^x = \log (3 - x)$$

$$x\log 3 = \log (3 - x)$$

$$\log 3 = \frac{\log (3 - x)}{x}$$

And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.
You may consider the function given by
$$
f(x)=3^x-(3-x),\quad x \in \mathbb{R}.
$$ We have
$$
f'(x)=3^x \cdot\ln3+1&gt;0,\quad x \in \mathbb{R},
$$ Thus the function is strictly increasing on $\mathbb{R}$.

We have
$$
\begin{align}
f(0)&amp;=1-(3-0)=-2&lt;0\\\\
f(1)&amp;=3-(3-1)=1&gt;0
\end{align}
$$ then the unique solution $x_0$ is such that $x_0 \in (0,1)$.

You may observe that
$$
3^x=3-x
$$ is equivalent to 
$$
(3-x)\ln 3 \times e^{(3-x)\ln 3}=3^3 \ln 3
$$ then, using a solution of $Xe^X=3^3 \ln 3$ in terms of the Lambert function, we get


  $$
x_0=3-\frac{W(27\ln 3)}{\ln 3}=\color{red}{0.741551813...}
$$
Edited version

After $x \cdot \log 3 = \log (3 – x)$ is $x = \frac {\log (3 – x)}{\log 3} $

which is equivalent to $y = x$ and $y = \frac {\log (3 – x)}{\log 3} $

Plotting the above TWO curves on the same graph, you will find they intersect at only one point (i.e. one solution).

Note also that:- When x > 3, the 'y' of the $y = x$ is much higher than the 'y' from $y = \frac {\log |3 – x|}{log 3}$. That means they will never meet when $x &gt; 3$.