0
$\begingroup$

I have a homework in real analysis and I'm very confused about it. I would be very thankful, if you could give me any ideas/tips or solutions how to get this task done. The task is as follows:

Prove that equation $$ 3^{x} + 4^x = 5^x $$ has exactly one real root. Hint that has been given: examine the function $f(x) = \frac{3^x}{5^x} + \frac{4^x}{5^x} - 1 $.

$\endgroup$
  • $\begingroup$ So $f(2)=0$. Now what can you say about the behavior of $f(x)$ around 2? $\endgroup$ – Laars Helenius Oct 27 '17 at 6:22
1
$\begingroup$

$$f'(x)=(3/5)^x \ln \frac35+(4/5)^x \ln \frac45;$$ $$ \ln \frac35<0,\;\;\ln \frac45<0 $$ $$ \forall x\in\mathbb R\quad (3/5)^x>0,\;\;(4/5)^x>0 $$ Therefore $\forall x\in\mathbb R\;\; f'(x)<0$, thus $f(x)$ is strictly decreasing $\Rightarrow$ $f(x)=0$ has only one root $x=2$ (because for $x>2\;$ $f(x)<f(2)=0$ and for $x<2\;$ $f(x)>f(2)=0$).

$\endgroup$
1
$\begingroup$

It is clear that $x=2$ is a solution.You have to show that there are no other solutions.The following solution uses only elementary inequality method(without calculus).

Note that $(3/5)^2+(4/5)^2=1$.Again for any given $x>2$ we have $(3/5)^x<(3/5)^2 ,(4/5)^x<(4/5)^2$ then $3^x+4^x<5^x$ for any given $x>2$.Now try yourself to prove similar useful inequality for $x<2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.