solve $e^x = x^{100}$ for $x$ Can't solve 
$$e^x = x^{100}$$
for $x$ to find where they intersect.
Setting them equal to each other and taking the $\ln$ of both sides gives $x = 100 \ln x$. And I'm stuck.  What am I missing?
 A: You can get solutions in terms of the Lambert W function, giving $$x = -100 W(-1/100)\\x = -100 W(1/100)\\x = -100 W_{(-1)}(-1/100)$$
or get numeric answers
$$x \approx 1.01015,-0.990147,647.278$$
One dimension root finding will find these.  It is discussed in any numerical analysis text.
A: You have $x = 100 \ln x$, which can be written $\frac{\ln x}{x} = 0.01$
Since this question is marked "precalculus", you will have to rely on qualitative arguments to see that  $\frac{\ln x}{x}$ will be postive for $x>0$, and tend monotonically towards zero for $x\to \infty$. Since  $\frac{\ln x}{x} =0$ at $x=1$, it's fairly clear that there is a peak value, which in fact occurs at $x=e$.

Thus the line $0.01$ will cut this curve twice in positive values. Numerical evaluation puts the two results near $x=1.01$ and $x=647$.
Since the exponent on $x$ is even, there is also a solution for negative $x$. This can be found similarly by considering $w=-x$ and  $e^{-w} = w^{100}$ and then looking for the point where the above curve cuts $-0.01,$ which is close to $w=0.99$.
A: Let's first suppose $x>0$. Then we can transform the equation into
$$
x=100\log x
$$
and we can consider the function $f(x)=x-100\log x$. Clearly
$$
\lim_{x\to0}f(x)=\lim_{x\to\infty}f(x)=\infty
$$
and we have
$$
f'(x)=1-100/x
$$
which only vanishes at $x=100$. Therefore we know that $x=100$ is a point of minimum for $f$. Now
$$
f(100)=100-100\log 100<0
$$
so we conclude that the equation $f(x)=0$ has two solutions.
If $x<0$, we can set $x=-y$ and the equation becomes $-y=100\log y$. If we consider $g(y)=y+100\log y$ we see that $g'(y)=1+100/y>0$, so the function is increasing. Since
$$
\lim_{y\to0}g(y)=-\infty,\qquad \lim_{y\to\infty}g(y)=\infty
$$
we conclude that the equation $g(y)=0$ has a single solution.
In total, the equation $e^x=x^{100}$ has three solutions.
A: Use the  Kantorovich's  theorem on Newton's
method in its classical formulation. Set the function $F:(-\infty,+\infty)\to \mathbb{R}$ by
$$
F(x)=e^x-x^{100}.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's  theorem on Newton's interactions. The method  sometimes (very rarely) results in a finite step method and thus exact solution.

Let $I\subseteq \mathbb{R}$ a interval
    and $F:{I}\to \mathbb{R}$ a continuous function, continuously
    differentiable on $\mathrm{int}(I)$. Take $x_0\in \mathrm{int}(I)$,
    $L,\, b>0$ and suppose that
  
  
*
  
*$F '(x_0)$ is non-singular,
  
*$ \|  F'(y)-F'(x)
    \| \leq L\|x-y\|
    \;\;$  for any $x,y\in I$,
  
*$ \|F'(x_0)^{-1}\cdot F(x_0)\|\leq b$,
  
*$2\cdot b\cdot L\leq 1$.
Define
    $
    t_*:=\frac{1-\sqrt{1-2bL}}{L}. 
  $
    If 
    $
  [x_0-t_*,x_0+t_*]\subset I,
  $
    then  the sequences $\{x_k\}$ generated by Newton's Method for
    solving $F(x)=0$ with starting point $x_0$,
  $$
    x_{k+1} ={x_k}-F'(x_k) ^{-1}F(x_k), \qquad k=0,1,\cdots, 
 $$
    is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
    point $x_*\in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
    $[x_0-t_*,x_0+t_*]$. 

Here a expository Article: Kantorovich's s theorem on  Newton's method.. See too other expository Article here.. See Ortega's paper here for a simple proof. 
