How do I show that $3^{x}+4^{x} = 5^{x}$ has exactly one real root.
Real roots of $3^{x} + 4^{x} = 5^{x}$
-
0Side note: complex solutions $x$ can also be studied. (Michael Lapidus has a book on "complex dimensions".) – 2012-06-04
3 Answers
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.
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 $\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \gt 1.$
-
1Ah nice. It's elementary – 2012-06-04
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.
-
0@GEdgar: you're right. OK. – 2012-06-04