I have the equation $f(x) = -n^{-x} × (n + 1)^{x - n - 2} × \left(\left((n + 1)^{n + 1} - n^n × x\right) × (\ln(n) - \ln(n + 1)) + n^n\right)$ where $n$ is a positive real number and $\ln(z)$ is the natural logarithm, and I want to solve for $f(x) = 0$ in terms of $n$.
Putting the equation into WolframAlpha gives the solution $x = \frac{1}{\ln(n) - \ln(n + 1)} - \frac{\operatorname{W}\left(-n^{-n} (n + 1)^{n + \frac{1}{\ln(n) - \ln(n + 1)}} \ln(n + 1)\right)}{\ln(n + 1)}$, where $\operatorname{W}(z)$ is the Lambert W function. However, when I plot that in GeoGebra, it doesn't line up with the point where $f(x)$ crosses the $x$-axis:
I'm not sure how to go about solving this by hand, with WolframAlpha apparently not working in this case.
Is there a closed-form solution to this equation? If so, how do I find it?
Your function $f$ contains three different factors; each of them may vanish. It is straightforward to see that the first two factors cannot be zero $-$ because $n^{-x} > 0$ for $n > 0$ and the same goes for the second exponential $-$, that is why you are left with the subequation $\left((n+1)^{n+1} - n^nx\right) \ln\left(\frac{n}{n+1}\right) = -n^n$, which is solved by $$ x = \frac{(n+1)^{n+1}}{n^n} + \frac{1}{\ln\left(\frac{n}{n+1}\right)} $$ after some simple algebraic manipulations. It is to be noted that $x \approx 11.16$ when $n = 6$, as observed on your graph. Not sure how WolframAlpha ended up with such an expression...