$\displaystyle {5^(2x-2x^2)}={24*5^-2}+{5^(-2x2)}$
Solve for X
That's my guess at what you really mean. That "$\displaystyle 24*5^{-2}$" really complicates things since we do NOT have just "$\displaystyle 5^a= 5^b$
We can rewrite this as $\displaystyle 5^{2x- 2x^2}- 5^{-2x^2}= 24*5^{-2}$ which is the same as $\displaystyle 5^{2x}5^{-2x2}- 5^{2x^2}= (5^{2x}- 1)(5^{-2x^2})= 24*5^{-2}$. Now, I note that if x= 1, then I have $\displaystyle 5^{2x}-1= 25- 1= 24$ and $\displaystyle 5^{-2x^2}= 5^{-2}$. However, in general, you cannot just assume that if $\displaystyle ab= 24*5^{-2}$ then $\displaystyle a= 24$ and $\displaystyle b= 5^{-2}$. We can see that x= 1 is a solution but we do not yet know if there are other solutions.
1. Re-write the equation:
$\displaystyle \dfrac{5^{2x}}{5^{2x^2}} = \dfrac{24}{25} + \dfrac{1}{5^{2x^2}}$
2. Collecting the variables at the LHS:
$\displaystyle \dfrac{5^{2x}}{5^{2x^2}} - \dfrac{1}{5^{2x^2}}= \dfrac{24}{25} $
$\displaystyle \dfrac{5^{2x}-1}{5^{2x^2}} = \dfrac{24}{25} $
3. That means you have to solve:
$\displaystyle 5^{2x}-1 = 24~\wedge~5^{2x^2} = 25$
$\displaystyle 5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2$
$\displaystyle 2x = 2~\wedge~2x^2=2$
which yields x = 1 as the only solution.
Thank you for pointing this out.
Unfortunately I was only able to get a approximate numerical solution:
$\displaystyle 5^{2x-2x^2}=24*5^{-2}+5^{-2x^2}$
$\displaystyle 5^{2x}=\frac{24}{25}\cdot 5^{2x^2}+1$
$\displaystyle 5^{2x}=24\cdot 5^{2x^2-2}+1$
$\displaystyle 0=24\cdot 5^{2x^2-2}-5^{2x}+1$
Now using Nerwton's method yields $\displaystyle x \approx 0.2381291343$
this does not follow The fact that $\displaystyle \frac{x}{y}= \frac{a}{b}$ does NOT imply that x= a and y= b!
It does happen to be true for x= 1 but it is not true that we must separate the fractions that way.
$\displaystyle 5^{2x} = 25 = 5^2~\wedge~5^{2x^2} = 25 = 5^2$
$\displaystyle 2x = 2~\wedge~2x^2=2$
which yields x = 1 as the only solution.