Solving $\log(x)P(x)=y$ where $P(x)$ is a polynomial How does one solve $\log(x)\cdot P(x)=y$, where $y$ is a positive constant, $P(x)$ is a given polynomial, and $\log$ can be $\log$ or $\ln$. Either numerically or exactly.
 A: One method to solve it numerically is to first convert the problem into a root-finding problem. Suppose that $\alpha \in (0, \infty)$ is a solution, then we have:
$$ f(x) := \log x \cdot P(x) - y,\\
f(\alpha) = 0.$$
Then certainly $f$ is arbitrarily often differentiable on its domain, $(0, 
\infty)$.
If $f'(\alpha) \neq 0,$ (which I'm not sure you can guarantee) we can use the Newton method to solve for $\alpha$:
$$ \phi(x):=x-\frac{f(x)}{f'(x)}.$$
Starting with a value $x_0$ (which may have to be picked experimentally), the iteration $x_{n+1} =\phi(x_n)$ should converge to the desired value. The picking of $x_0$ will determine this, though; pick a wrong one and it'll end up diverging, rapidly.
Note: if $f'(\alpha) =0$, then the iteration $x_{n+1} = f(x_n)$ should do the trick as well; the fact that the function being iterated has a derivative of $0$ at the point we're looking to find speeds up the iteration (or, makes it diverge faster), provided the function is smooth enough ($C^2$ in this case).
A: In general, you can't explicitly. However, if $P(x)=\alpha x^\beta$, set $z:= x^\beta$ and $t:=\log z$ then
$$
y=\log x P(x)=\alpha \log x x^\beta=\frac{\alpha}{\beta} z \log z \implies \frac{\beta}{\alpha}y=te^t.
$$
Therefore
$$
t=W\left(\frac{\beta}{\alpha}y\right) \implies z=e^{W\left(\frac{\beta}{\alpha}y\right)} \implies x=e^{\frac{1}{\beta}W\left(\frac{\beta}{\alpha}y\right)},
$$
where $W$ is the Lambert W function.
