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Show that the equation $\displaystyle 3^x+4^x=5^x$ has exactly one root

I proved it graphically but I am in need of an analytic solution

prat
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5 Answers5

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HINT:

Observe that $\frac35,\frac45<1,$

So, $$\left(\frac35\right)^m+\left(\frac45\right)^m<\left(\frac35\right)^2+\left(\frac45\right)^2=1 \text{ for }m>2$$

and $$\left(\frac35\right)^m+\left(\frac45\right)^m>\left(\frac35\right)^2+\left(\frac45\right)^2=1 \text{ for }m<2$$

This can be generalized for any Pythagorean triple

lab bhattacharjee
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$3^x+4^x=5^x\iff\left(\frac35\right)^x+\left(\frac45\right)^x=1$. Since the exponential function on the left hand side is strictly decreasing, it follows that the equation, were it to ever be true for a certain x, then that x is unique. And it is indeed true for $x=2$, making it its only solution.

Lucian
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A hint: Use that $$x\mapsto\left({3\over5}\right)^x+\left({4\over5}\right)^x=e^{-\log\!{5\over3}\>x}+e^{-\log\!{5\over4}\>x}$$ is strictly decreasing on ${\mathbb R}$.

Christian Blatter
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Hint: Use the Intermediate Value Theorem to show that $f(x) = 3^x+4^x-5^x$ has a zero on some interval. The use Rolle's theorem to show you can't have more than one zero.

Christopher Toni
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  • I.V.T proves there is a zero $x=a$ between 1 and 3. Now I assume there is a zero at $x=b$. So that there is a $x=d$ between a and b such that $f'(d)=0$. How do I show now that there is no value which can satisfy this? – prat Nov 13 '13 at 09:30
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enter image description here write 5^x=(4+1)^x and use binomial expansion

then simplify

it seems to have answer only for x=2

Khosrotash
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