Find integer $x$ so that $$5\cdot5^{2x}-24\cdot2^x - 5 = 0$$
I know there is an approach as we can show the function $f(x):= 5\cdot5^{2x}-24\cdot2^x - 5$ is injective.
So, the equation can have at most one solution. And, basically, if we can guess the solution, that's it.
Is there any other way to resolve this equation? Here is what I've tried:
$$5\cdot(5^{2x} - 1)=24\cdot2^x$$
The last digit of the left member can be $0$, and the last digit of the right member can be $2$, $4$, $6$ or $8$. Therefore, there is no solution in $\mathbb Z$.
This equation has not solution for $x\ge 0$, since $5\not| 24\cdot 2^x$. For $x<0$, we define $y=-x$ and write the equation as $$ {5(25^{-y}-1)=24\cdot 2^{-y}\implies \\ 5\cdot 2^y\cdot (1-25^y)=24\cdot 25^y\implies \\ 2^y\cdot (1-25^y)=24\cdot 5^{2y-1}, } $$which, again, has no solution for $y\ge 1$ since $5\not | 2^y\cdot(1-25^y)$ and the equation has no integer solution.