Solve in $\mathbb R\quad 5^\sqrt{x} - 5^{x-7} = 100$ Solve in $\mathbb R$
$$5^\sqrt{x} - 5^{x-7} = 100$$
$\mathbf {My Attempt}$
I converted the eq. to this form
$$5^{(\sqrt{x}-3)(\sqrt{x}+3)}-5.5^{\sqrt{x}-3}+4=0$$
It's apparent that $\;\mathbf {x=9}\; $ is a solution, but I can't find the reasoning for this mathematically.
Any hint?
$\mathbf {Edit}$
I'll post my solution tomorrow.
 A: My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)\to \mathbb{R}$ by
$$
F(x)=5^x-\frac{1}{5^7}5^{x^2}-100.
$$
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.
Use the  Kantorovich's  theorem on Newton's
method in its classical formulation. 

Let $I\subseteq \mathbb{R}$
    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
.1 $F '(x_0)$ is non-singular,
.2 $ \| F'(x_0)^{-1}\left[ F'(y)-F'(x)\right]
    \| \leq L\|x-y\|
    \;\;$  for any $x,y\in I$,
.3$ \|F'(x_0)^{-1}F(x_0)\|\leq b$,
.4 $2bL\leq 1$.
Define
    \begin{equation}
    t_*:=\frac{1-\sqrt{1-2bL}}{L},\qquad
    t_{**}:=\frac{1+\sqrt{1-2bL}}{L}. 
  \end{equation}
    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 $x_0$,
    \begin{equation} \label{ns.KT} 
    x_{k+1} ={x_k}-F'(x_k) ^{-1}F(x_k), \qquad k=0,1,\cdots, 
  \end{equation}
    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
    $[x_0-t_*,x_0+t_*]$ and 
    \begin{equation}
    \label{eq:q.conv.x}
    \|x_*-x_{k+1}\|\leq \frac{1}{2} \|x_*-x_k \|, \qquad
    k=0,1,\,\cdots. 
  \end{equation}
     Moreover, if assumption .4 holds as an strict inequality, i.e.
     $2bL<1$, then
     \begin{equation} 
     \|x_*-x_{k+1}\|\leq\frac{1-\theta^{2^k}}{1+\theta^{2^k}}
     \frac{ L}{2\sqrt{1-2bL}}\|x_*-x_k\|^2\leq
     \frac{ L}{2\sqrt{1-2bL}}\|x_*-x_k\|^2, \quad k=0,1,\cdots, 
   \end{equation}
    where $\theta:=t_*/t_{**}<1$, and $x_*$ is the
    unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $\rho$ such that
    $ t_*\leq\rho<t_{**},\qquad [x_0-\rho,x_0+\rho]\subset I.$

A: Note: The right-hand side $100$ has a nice representation by powers of $5$.

We have
  \begin{align*}
5^{\color{blue}{\sqrt{x}}}-5^{x-7}=100=5^\color{blue}{3}-5^2
\end{align*}
  which indicates a trial via
  \begin{align*}
\sqrt{x}=3 \qquad\text{and}\qquad x-7=2
\end{align*}
  giving the solution $x=9$.

I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
A: This is the method I used to justify my found solution
Let  $\; t=5^{\sqrt{x}-3}\;$, we arrive at
$$t^{\sqrt{x}+3}-5 t+4=0$$ 
$$t^{\sqrt{x}+3} - t^2 + t^2 -5 t+4=0$$ 
$$t^2(t^{\sqrt{x}+1} -1)+(t-1)(t-4)=0$$
Assume $\;\sqrt{x} \in \mathbb N$
$$t^2(t-1)(t^{\sqrt{x}}+^{\sqrt{x}-1}+\cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^{\sqrt{x}+2}+^{\sqrt{x}+1}+\cdots +t^2+t-4)=0$$
$$t-1=0 \quad \Rightarrow\quad 5^{\sqrt{x}-3}=1 \quad \Rightarrow\quad \sqrt{x}=3 \in \mathbb N$$
$x=9\quad $ is a valid solution.
