# Real roots of $3^{x} + 4^{x} = 5^{x}$

How do I show that $3^{x}+4^{x} = 5^{x}$ has exactly one real root.

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Side note: complex solutions $x$ can also be studied. (Michael Lapidus has a book on "complex dimensions".) –  GEdgar Jun 4 '12 at 13:12

Hints:

• Let $\displaystyle f(x) = \biggl(\frac{3}{5}\biggr)^{x} + \biggl(\frac{4}{5}\biggr)^{x} -1$

• Note : $f'(x) < 0$ and $f(2)=0$. Apply Rolle's Theorem.

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Rewrite our equation as $$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1.$$ We have the familiar solution $x=2$.

If $x>2$, then $\left(\frac{3}{5}\right)^x \lt \left(\frac{3}{5}\right)^2$ and $\left(\frac{4}{5}\right)^x \lt \left(\frac{4}{5}\right)^2$, and therefore $$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \lt 1.$$ Similarly, if $x<2$ then $\left(\frac{3}{5}\right)^x \gt \left(\frac{3}{5}\right)^2$ and $\left(\frac{4}{5}\right)^x \gt \left(\frac{4}{5}\right)^2$, so $$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \gt 1.$$

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Ah nice. It's elementary –  Aaron Jun 4 '12 at 6:03

Consider our equation as: $$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1.$$

Notice that left side is a sum of 2 monotonically decreasing functions and their sum is a monotonically decreasing function. Hence, the only possible solution is x=2.

The proof is complete.

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Do not say "nonincreasing", since there could be multiple solutions for nonincreasing function in general. Instead say "decreasing". –  GEdgar Jun 4 '12 at 13:09
@GEdgar: you're right. OK. –  Chris's sis Jun 4 '12 at 13:14