TITLE: Trying to solve $x^2=2^x$ : What's wrong below?
QUESTION [2 upvotes]: I just learned about Lambert W Function and I am trying to solve the above equation (and I have seen this article but I want to solve it myself). However, I seem not to be able to reach the correct answer. Can someone please point out the error ?
$$x^2 = 2^x$$
Taking $\log_e$ (denoted by $\ln$) on both sides
$$\ln(x^2) = \ln(2^x)$$
$$\implies 2\;\ln(x) = x(\ln 2)$$
$$\implies \ln(x) = x \;(\ln 2)/2$$
$$\implies 1/(x\ln(x)) = (\ln 2)/2$$
Let $y = 1/x$, then we get
$$y\;\ln (1/y) = (\ln 2)/2$$
$$\implies y\;\ln y^{-1} = (\ln 2)/2$$
$$\implies -y\;\ln y = (\ln 2)/2$$
$$\implies y\;\ln y = -(\ln 2)/2$$
Replacing $y$ by $e^{\ln y}$, we get
$$\implies e ^ {\ln y}\;\ln y =  -(\ln 2) / 2$$
$$\implies \ln\;y = W(-(\ln 2)/2)$$
Replacing back $y = 1/x = x^{-1}$, we get
$$\implies \ln x^{-1} = W(-(\ln 2) / 2)$$
$$\implies \ln x = - W(-(\ln 2) / 2)$$
$$\implies x = e^{- W(-(\ln 2) / 2)}$$
And this isn't the right answer :(

REPLY [2 votes]: The function
$$
f(x)=\frac{\ln x}{x},
$$
is strictly increasing is $(0,e]$ and strictly decreasing in $[e,\infty)$.
since $f'(x)=\frac{1-\ln x}{x^2}$. Hence, for every $y\in (0,1)$ is the image of exactly two $x$'s. In particular, for every $y\in (0,1)$, there exist a unique $x_1\in (1,e)$ and a unique $x_2\in (e,\infty),$ such that
$$
y=\frac{\ln x_1}{x_1}=\frac{\ln x_2}{x_2}.
$$
Observe that
$$
\frac{\ln 2}{2}=\frac{\ln 4}{4},
$$
and $2<e<4$. Hence the only solutions of
$$
2^x=x^2 \quad\Longleftrightarrow\quad \frac{\ln x}{x}=\frac{\ln 2}{2},
$$
are 2 and 4.
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