Question

find the value or values of x that can satisfy the equation.If anybody have dare to solve the solve it​

Answer 1

$\begin{aligned}&\sqrt[x-4]{5^{\dfrac{x}{\sqrt x+2}}\cdot(0.2)^{-\dfrac{4}{\sqrt x+2}}}=125\cdot(0.04)^{\dfrac{x-2}{x-4}}\\\\\implies\ \ &\left({5^{\dfrac{x}{\sqrt x+2}}\cdot\left(\dfrac{1}{5}\right)^{-\dfrac{4}{\sqrt x+2}}\right)^{\dfrac{1}{x-4}}=5^3\cdot\left(\dfrac{1}{25}\right)^{\dfrac{x-2}{x-4}}\\\\\implies\ \ &\left({5^{\dfrac{x}{\sqrt x+2}}\cdot5^{\dfrac{4}{\sqrt x+2}}\right)^{\dfrac{1}{x-4}}=5^3\cdot5^{\dfrac{-2(x-2)}{x-4}}\end{aligned}$

$\begin{aligned}\left({5^{\dfrac{x+4}{\sqrt x+2}}\right)^{\dfrac{1}{x-4}}=\left(5^{3(x-4)-2(x-2)}\right)^{\dfrac{1}{x-4}}\end{aligned}$

So we get,  

$\begin{aligned}&\dfrac{x+4}{\sqrt x+2}=3(x-4)-2(x-2)}\\\\\implies\ \ &\dfrac{x+4}{\sqrt x+2}=x-8\\\\\implies\ \ &x+(x-8)\sqrt x=20\end{aligned}$

Now we put some logic. Well, nothing, we give some values for x, whether the LHS of this equation comes closer to 20.

$x+(x-8)\sqrt x\quad\longrightarrow\quad(1)$

One thing is sure is that x is non-negative.

$x=0\quad\implies\quad(1)=0\\\\x=1\quad\implies\quad(1)=-6\\\\x=2\quad\implies\quad(1)\to-4$

Likewise, skipping some values, finally,

$x=9\quad\implies\quad(1)=12\\\\x=10\quad\implies\quad(1)\approx16\\\\x=11\quad\implies\quad(1)\approx20$

Now, (1) also increases as x increases.

(1) will be 20 for a value of x where 10 < x < 11. No other case can't be found.

Hence there's only 1 value for x which satisfies the condition.