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.
 A: 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.
A: 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.
A: 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
$$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \gt 1.$$
