There are a few problems with addition.
$$W(x+a)=?$$
$$\log(x+a)=?$$
The inverse functions of exponential related functions don't like addition on the inside. Addition on the outside is perfectly fine though.
$$\log(x)+a=\log(e^ax)$$
Except even then, the Lambert W Function won't simplify if you have addition on the outside... most of the time. And this question isn't one of the ones that simplifies.
So the best is to go with linear approximation methods.
This is just a partial answer.
Since I faced this specific problem a few years ago for $W_0(x)$, what I found as "best" are Padé approximants.
The simplest ones are
For $\color{red}{-\frac 1e \leq x \leq -\frac 1{2e}}$
$$W_0(x) \approx \frac{-1+\frac{14\sqrt{2}}{45} \sqrt{e x+1}+\frac{301}{540} (e x+1)}{1+\frac{31\sqrt{2}}{45} \sqrt{e x+1}+\frac{83}{540}
(e x+1)}\tag 1$$
For $\color{red}{-\frac 1{2e} \leq x \leq 0}$
$$W_0(x) \approx \frac{x+\frac{4 }{3}x^2}{1+\frac{7 }{3}x+\frac{5 }{6}x^2}\tag 2$$
For sure, for more accuracy, I built similar expressions with more terms.
Edit
For the other branch, you can use
$$W_{-1}(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(-2+L_2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\frac{L_2(-12+36L_2-22L_2^2+3L_2^3)}{12L_1^4}+\frac{L_2(60-300L_2+350L_2^2-125 L_2^3+12 L_2^4)}{60L_1^5}+\cdots$$ where $L_1=\log(-x)$ and $L_2=\log(-L1)$
Update
Calling $f(x)$ and $g(x)$ the expressions given in $(1)$ and $(2)$, let us consider the error function
$$\Phi(a)=\int_{-\frac 1 e}^a \left(W(x)-f(x)\right)^2+\int^0_a \left(W(x)-g(x)\right)^2$$ and get the following values
$$\left(
\begin{array}{cc}
a & 10 ^{10} \,\Phi(a) \\
-0.350 & 472433 \\
-0.325 & 72500 \\
-0.300 & 13455 \\
-0.275 & 2646 \\
-0.250 & 523 \\
-0.225 & 109 \\
\color{red}{ -0.200} &\color{red}{ 44} \\
-0.175 & 57 \\
-0.150 & 100 \\
-0.125 & 168 \\
-0.100 & 269 \\
-0.075 & 411 \\
-0.050 & 605 \\
-0.025 & 863 \\
0.000 & 1197
\end{array}
\right)$$
Best Answer
Answer to your explicit question :
Observe that ,
$$ \begin{align}W\left(3^4\ln 3\right)&=W\left(3^3\cdot 3\ln 3\right)\\ &=W\left(3^3\ln 3^3\right)\\ &=W\left(\color{#c00}{\ln 3^3}\cdot e^{\color{#c00}{\ln 3^3}}\right)\\ &=\ln 3^3=3\ln 3\end{align} $$
Therefore, you obtain that :
$$ \begin{align}x&=4-\frac{W\left(3^4\ln 3\right)}{\ln 3}\\ &=4-\frac {3\ln 3}{\ln 3}\\ &=4-3=1\thinspace .\end{align} $$
$\rm{Construction :}$
Substituting $\thinspace 3^x=u\thinspace $, you get $\thinspace x=4-u\thinspace $ or $\thinspace x=\dfrac {\ln u}{\ln 3}\thinspace $, which leads to the following :
$$ \begin{align}x+3^{x}&=4\\ \frac{\ln u}{\ln 3}+u&=4\\ \ln u+u\ln 3&=4\ln 3\\ \ln \left(u\cdot e^{u\ln 3}\right)&=4\ln 3\\ u\ln 3\cdot e^{u\ln 3}&=e^{4\ln 3}\cdot \ln 3\\ u\ln 3&=W\left(3^4\ln 3\right)\end{align} $$
Then, the exact value becomes :
$$x=4-\frac{W\left(3^4\ln 3\right)}{\ln 3}\thinspace .$$