Solve in $\mathbb R\quad 5^\sqrt{x} - 5^{x-7} = 100$

Question

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.

Answer 1

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.$

Answer 2

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.

Answer 3

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.