Proof using the Lambert W function that 1 = 0 - What went wrong?

Question

All values that satisfy $x^2=2^x$ would satisfy $\ln(2)x^3 = x\ln(2)e^{x\ln(2)}$, and would therefore satisfy the relationship $W(\ln(2)x^3) = x\ln(2)$. The problem is that when I graph these functions, there are four real solutions to this relation: $x \approx -0.77$, and $x=4, 2, 0$. The first three are solutions to $2^x=x^2$, but $0$ is not. So what happened?

Answer 1

$$ x^2 =2^x $$ $$ 2\ln(x)=x\ln(2) $$ Note that $x \neq 0 $, since $\lim_{x\to0}\ln(x) = -\infty$

So you accidentally introduced a extraneous solution. It is much like the classic proof:

$$ a =b $$ $$a^2 = ab$$ $$a^2-b^2 = ab - b^2$$ $$ (a-b)(a+b) = b(a-b)$$ $$a+b = b$$ But $b=a$ $$2b = b$$ $$2=1$$ $$\forall a \& b \neq 0$$

With the mistake being that we can't divide by $(a-b)$, since $a-b = 0$

Answer 2

Let's use your technique to solve $3x=6$. We will first multiply by $x$ [this is the "mistake"] to get $3x^2=6x$. Subtracting and factoring gives $3x(x-2)=0$, which has the two solutions $x=0$ and $x=2$. Of these, only the second is a solution to our original equation. The "solution" $x=0$ is extraneous/false because multiplying any equation $A=B$ at all by $0$ on both sides results in something true (namely $0=0$) whether or not $A=B$ is true.