How to solve $x^2 - 4^x = 0$? Solving the following equation
$$x^{100} - 4^xx^{98} - x^2 + 4^x = 0$$
yielded
$$(x^2 - 4^x)(x^{98} - 1) = 0$$
From the right term on the left side, I get the solutions $x = \pm 1$, but I'm unsure how to solve the left term on the left side without using Wolfram. Is it possible? Thanks in advance.
 A: Take the square root of both sides to get
$$x=\pm2^x$$
We can see that
$$x=2^x$$
has no solutions, since for $x\le0$, we have
$$x\le0<2^x$$
And for $x>0$, we have
$$2^x>x\ln(2)>x$$
However, it is easy to see that
$$x=-2^x$$
has solutions, since
$$f(x)=x+2^x$$
is continuous and $f(-1)<0<f(0)$. This solution may be found using the Lambert W function, as follows:
$$x=-2^x=-e^{x\ln(2)}$$
$$-x\ln(2)e^{-x\ln(2)}=\ln(2)$$
$$-x\ln(2)=W(\ln(2))$$

$$x=-\frac{W(\ln(2))}{\ln(2)}=-e^{-W(\ln(2))}\approx−0.641185744505$$

To calculate this root, we may implement an Euler-like iteration method, noticing that $(-2^x)'<0$.
$$a_n=\begin{cases}t,&n=0\\\cfrac{a_{n-1}-2^{a_{n-1}+1}}3,&n>0\end{cases}$$
For any $t\in\Bbb[-1,0]$, we have,
$$x=\lim_{n\to\infty}a_n$$
For example, with $t=0$, we have
$$a_8\approx-0.641185744508$$
There are various ways to improve this method. By properly weighting terms, we can derive an even faster converging sequence:
$$b_n=\begin{cases}t,&n=0\\\cfrac{b_{n-1}^2\ln(2)+2^{b_{n-1}}}{b_{n-1}\ln(2)-1},&n>0\end{cases}$$
With $t=0$, we get
$$b_5\approx-0.641185744511$$
Asymptotically, when $t=0$, we have
$$a_n=x-\mathcal O\big(0.03705^n\big)$$
$$b_n=x-\mathcal O\big(|0.5865x|^{2^n}\big)$$
(assuming I did that correctly)
Notice how quickly $b_n$ converges compared to $a_n$!
A: $$x^2 - 4^x = 0\Longleftrightarrow \ln |x| = x \ln 2\Longleftrightarrow f(x)= 0 $$
Where $f(x) =\ln | x| -x\ln 2$
$$f'(x) =  \frac{1}{x}-\ln 2=0\implies x= \frac{1}{\ln 2} $$
But $$\lim_{x\to 0^+} f(x) = \lim_{x\to \infty} f(x) = -\infty$$
Hence on $(0,\infty)$ $f$ reach its maximum at $\frac{1}{\ln 2}$ but 
$$f\left(\frac{1}{\ln 2}\right) = \ln \frac{1}{\ln 2} -1<0 \implies f(x)<0 \forall~~,x>0$$
So the equation $f(x) = 0$ has no solution on $(0,\infty)$
But on  $(-\infty,0)$ we have that $f$ is strictly decreasing and
$$\lim_{x\to 0^-} f(x) = -\infty\quad\text{and}\quad\lim_{x\to -\infty} f(x) = \infty$$
Therefore the equation $f(x)=0$ has only one  solutions $x_1$ such that:
 $$-1<x_1< -\frac{1}{2}.$$
Since $$f(-1) = \ln2>0\quad\text{and}\quad f(-\frac{1}{2}) =-\frac{1}{2}\ln 2<0$$
