<s>[INST] How should I check that $6^{\log_{10} x} + 8^{\log_{10} x} = x $ does not have other solutions? The question is to solve the equation $6^{\log_{10} x} + 8^{\log_{10} x} = x $
I know one of the solutions is $x=100$ using Pythagorean triples but I can't show that this is the only solution.
I'm looking for an idea without using differentiation!
any hints would be appreciated. tnx
 [/INST] Solving for $x$ in $6^{\log_{10} x} + 8^{\log_{10} x} = x$
is equivalent to solving for $y$ in $6^y + 8^y = 10^y$ where $y = \log_{10} x.$
I have an easier time visualizing this equation with $y$ than the one with $x,$ so I would prefer to solve for $y$.
You can also write this $6^{y-2} 6^2 + 8^{y-2} 8^2 = 10^{y-2} 10^2.$
You know you have a solution for $y = 2,$ because $6^2 + 8^2 = 10^2.$
This means you can also write other statements that are always true, such as
$6^{y-2} 6^2 + 6^{y-2} 8^2 = 6^{y-2} 10^2$
or $8^{y-2} 6^2 + 8^{y-2} 8^2 = 8^{y-2} 10^2.$
For $y > 2,$ you have $6^{y-2} < 8^{y-2} < 10^{y-2}.$
For $y < 2,$ you have  $6^{y-2} > 8^{y-2} > 10^{y-2}.$
Now you just need to find a way to compare $6^{y-2} 6^2 + 8^{y-2} 8^2$
to $10^{y-2} 10^2$ in each of those two cases.
It might even help to make a further substitution, such as $t = y - 2,$
so now you're trying to solve for $t$ in $6^t 6^2 + 8^t 8^2 = 10^t 10^2,$
you know $t=0$ is a solution, and you want to show there are no solutions for
$t > 0$ or for $t < 0.$  Just like it is easier to think about $6^y$ than
$6^{\log_{10} x}$ (at least for me; your mileage may vary),
it may be easier to think about $6^t$ than $6^{y-2}.$
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