Is this a good way to prove that $3^x+4^x =5^x $ has $x=2$ as the only real solution?

Question

Divide both sides of the equation $3^x+4^x=5^x$ by $5^x$.

$$ \Rightarrow \frac { 3^x }{ 5^x } +\frac { 4^x }{ 5^x } =\frac { 5^x }{ 5^x }$$ $$\tag1 \Rightarrow \left( \frac 3 5 \right)^x + \left( \frac { 4 }{ 5 } \right)^x =1$$ $\because \frac { 3 }{ 5 } \leqslant 1 \Rightarrow \sin \theta =\frac { 3 }{ 5 }$ would be valid.

We know, $\sin^2 \theta +\cos ^2 \theta =1$.

$\Leftrightarrow \cos^2 \theta =1-\sin ^2 \theta$.

$$ \Leftrightarrow \cos \theta =\sqrt { 1- \left( \frac { 3 }{ 5 } \right)^2 } =\frac { 4 }{ 5 }$$

$\therefore \sin \theta =\frac { 3 }{ 5 } \ \land \ \cos\theta =\frac { 4 }{ 5 }$.

The equation $(1)$ can be rewritten as $$ \Rightarrow \left( \cos\theta \right)^x + \left( \sin \theta \right)^x =1.$$

We know that the above equation would hold true for only $ x=2$.

Hence Proved.

Thanks to lab bhattacharjee for reminding me to link the question details to other answers.

Please tell me if my answere is better and more reasonable than the other answers on the following pages:

Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$

Proving that $ 2 $ is the only real solution of $ 3^x+4^x=5^x $

If it is not the best one, please give the link to the best solution. Thanks a ton for your time!

Thanks to Hagen von Eitzen for reformatting the question details into a better format.

Answer 1

Let $f(x)=(\frac{3}{5})^x+(\frac{4}{5})^x$

This function is strictly decreasing , and as such it's graph will only intersect the line $y=1$ in only one point(suppose there are two points, due to the fact that it is strictly decreasing, these points must coincide).