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
 A: $$\frac{d}{dx} (6^{\log_{10} x} + 8^{\log_{10} x}) = \frac{\log (6) 6^{\frac{\log (x)}{\log (10)}}+\log (8) 2^{\frac{3 \log (x)}{\log (10)}}}{x \log
   (10)} < 1$$
for $x >100$:

A: \begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
    &= 6^{\log_6(x)/ \log_6 10} + 8^{\log_8(x)/ \log_8 10}  \\
    &= x^{1/\log_6 10} + x^{1/\log_8 10}  \text{.}
\end{align*}
We have
\begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
    &= x  \qquad \text{when}  \\
\frac{6^{\log_{10} x} + 8^{\log_{10} x}}{x} &= 1    \qquad \text{ so when}  \\
\frac{x^{1/\log_6 10} + x^{1/\log_8 10}}{x} &= x^{-1 +1/\log_6 10} + x^{-1 + 1/\log_8 10}  \\
   &= 1  \text{.}
\end{align*}
We know this happens when $x = 100$.  Also,
$$  \frac{\mathrm{d}}{\mathrm{d}x} \left( x^{-1 +1/\log_6 10} + x^{-1 + 1/\log_8 10} \right) = \left( -1 +1/\log_6 10  \right) x^{-2 +1/\log_6 10} + \left( -1 + 1/\log_8 10 \right) x^{-2 + 1/\log_8 10}  \text{.}  $$
Recall the original equation, in which the domain of $\log_{10} x$ is $x > 0$.  For $x > 0$, the powers of $x$ in this derivative are positive.  Also, $-1 +1/\log_6 10 = -0.22{\dots} < 0$ and $-1 +1/\log_8 10 = -0.09{\dots} < 0$, so the derivative is negative on the entire domain.  Therefore, $\displaystyle \frac{6^{\log_{10} x} + 8^{\log_{10} x}}{x}$ is strictly monotonically decreasing, so can only take the value $1$ at most once.
We can even determine that $-1 +1/\log_6 10 < 0$ without a calculator.  Since $10 > 6$, $\log_6 10 > 1$ and taking reciprocals, $\frac{1}{\log_6 10} < \frac{1}{1} = 1$, so we don't add enough to $-1$ to get a positive result.  A parallel argument shows the same thing about $-1 +1/\log_8 10$.
A: 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 $6^{y-2}.$
