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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.

Mr Pie
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Prabal Gupta
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    http://math.stackexchange.com/questions/61812/proving-that-2-is-the-only-real-solution-of-3x4x-5x – lab bhattacharjee Apr 29 '14 at 16:44
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    @labbhattacharjee Sorry, but the question was if my solution was more reasonable than the one's you just provided the link for. Why or Why not? – Prabal Gupta Apr 29 '14 at 16:46
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    At the end there is the assertion that $x=2$ is the only solution of $(\cos \theta)^x+(\sin\theta)^x=1$. This needs to be proved. The introduction of sine and cosine is not really necessary, but is fine if it helps you to visualize. – André Nicolas Apr 29 '14 at 16:50
  • You should only put mathÜ in Latex markup, not everything*. Legibility really suffers. – Hagen von Eitzen Apr 29 '14 at 16:52
  • @HagenvonEitzen Sorry, My fault. I will change it soon as possible. Thanks for the suggestion! – Prabal Gupta Apr 29 '14 at 16:53
  • @HagenvonEitzen I am grateful to you as you undertook effort to correct my mess. Thanks! – Prabal Gupta Apr 29 '14 at 16:58
  • @AndréNicolas, Sir, isn't ${ \left( \cos { \theta } \right) }^{ x }+{ \left( \sin { \theta } \right) }^{ x }=1$ an identity true for only x=2? I am not really a good student, it'd be nice if you could explain further. – Prabal Gupta Apr 29 '14 at 17:03
  • You need to prove it, that is basically what the problem is about, since verifying that $3^2+4^2=5^2$ is easy. Rewrite as $(3/5)^x+(4/5)^x=1$, use the fact that (for say positive $x$ and $y$) and $0\lt s\lt 1$, if $x\lt y$ then s^x\gt s^y$. – André Nicolas Apr 29 '14 at 17:25
  • Or even more intuitive: Consider the form $(3/4)^x+1=(5/4)^x$. Then the left side is decreasing and the right increasing. – Lutz Lehmann Apr 29 '14 at 18:00
  • Note you are talking about exponential functions not trigonometric functions. The formula cos^2+sin^2=1 requires the number 2. Therefore since x is variable it is not possible to take advantage of that formula. –  Apr 30 '14 at 16:24
  • 5 to the power of x divided by 5 to the power of x is always 1. Clearly 5 to the power of x is not 0. (3/5)^x is strictly decreasing, so is (4/5)^x. Since this function is bijective there can only be one solution. –  Apr 30 '14 at 16:13

1 Answers1

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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).

shooting-squirrel
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