Algebraic solution to $2x + \ln x=2$ The answer to $2x+\ln x=2$ is $x=1$, which can be easily guessed after a failed attempt at solving it. I've tried to manipulate this into some inverse function of $2x+\ln x$, more specifically if $f(x)=2x+\ln x$, then I've tried to find $f^{-1}(x)$. But all I've gotten to is $xe^{2x}=e^2$, and I'm not sure how to simplify further.
Is there an algebraic solution to this? It seems like the product logarithm might be used, but I've only seen that function from Wolfram Alpha, and I'm not entirely sure how to use it.
Reiterating comments and complaints.
The way I read your question, you are asking for some function or method to algebraically/analytically solve the equation.  If we were told for instance that $x^2=4$ and were asked to find the positive value of $x$ satisfying this, sure we could say &quot;Oh, $2$ is the answer because look here... $2^2$ does in fact equal $4$&quot; but if we were to be asked about a similar question such as $x^2=3$ we would be hard-pressed to say &quot;Oh, $1.732050807568877\dots$ is the answer because $(1.732050807568877\dots)^2=3$.&quot;
For the question of solving $x^2=3$ for $x$, we were able to introduce a notation and concept... the square root... which allows us to answer this easily.  $x^2=3\implies x=\sqrt{3}$.  The way I read your question, you are looking for the analogous &quot;operation&quot; which lets us extract the value of $x$ from the above equation which uses this operation rather than just spotting the solution.  Spotting the solution for a particular case and proving it is indeed the solution is all well and good however it does nothing to help us find a solution in a similar problem with slightly changed values.
The answer to whether or not there is such an operation to solve for $x$ is that hoping for such an operation which solves a transcendental equation such as this is futile.  Any such operation will only be as useful as saying the solution is &quot;the solution&quot; and will be non-elementary.  In some related problems you will not have any conveniently named function to use.  In other problems you may have a function such as the Lambert-W function (a.k.a. the product-log function as you alluded to in your post) but actually using the Lambert-W in practice will generally require numerical methods to evaluate.
For this specific problem, looking at the generalized equation of $2x+\ln(x)=a$ you can indeed use the Lambert-W function.  Modifying the equation a bit, we have $e^{2x+\ln(x)}=e^a$ and $xe^{2x}=e^a$ and so $(2x)e^{(2x)}=2e^a$
This is of the form $we^w=z$ whose solution is $w=W_k(z)$ for some integer $k$ where $W_k$ is the Lambert-W function on the $k$'th branch.  That is, replacing the values of $w$ and $z$ for what we had $2x = W_k(2e^a)$ giving us
$$2x+\ln(x)=a\iff x = \frac{W_k(2e^a)}{2}$$
In the specific case of $a=2$ this does simplify to $W_k(2e^2)/2$ which simplifies to $1$, again using that $we^w=z\iff w=W_k(z)$
In the specific case of $a=3$ (so the equation $2x+\ln(x)=3$) this would not simplify well and would just be written as $W_k(2e^3)/2$ which one can find via numerical methods is approximately $1.3499618380\dots$
As noticed in the comments, in general we can't find elementary algebraic solutions to this kind of equations.
To proceed by elementary inequalities, in this special case we have that
$$2x + \ln x=2\iff \ln (x)=2(1-x)$$
and since

for $x&lt;1 \quad \ln (x)&lt;0 \,\land\, 2(1-x)&gt;0$
for $x&gt;1 \quad \ln (x)&gt;0 \,\land\, 2(1-x)&lt;0$
for $x=1 \quad \ln (x)=2(1-x)=0$

$x=1$ is the unique solution for the given equation.