# Roots of $f(x)=3^x+4^x-5^x$ [duplicate]

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Let define $$f:\mathbb{R}\rightarrow\mathbb{R}$$ as $$f(x)=3^x+4^x-5^x$$.

Prove that there exists only one $$x_0$$ such as $$f(x_0)=0$$.

My approach:

We can see that $$\lim_{x\rightarrow-\infty}f(x)=0$$ and $$f(2)=0$$. I don't know how to use the derivatives of $$f$$ to prove that it is firstly increasing and then decreasing for $$x\in(-\infty,2)$$ and for $$x\in(2,\infty)$$ it is only decreasing.

## marked as duplicate by Dietrich Burde, Thomas Shelby, Javi, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 20 at 16:16

For $$x>2$$ we have $$3^{x}+4^{x}=(9)3^{x-2}+(16)4^{x-2}<(9)5^{x-2}+(16)5^{x-2}=5^{x}$$ and the inequalities gets reversed for $$x<2$$. Hence $$x=2$$ is the only solution.
Easier way to see it is $$3^x+4^x=5^x \qquad\Longleftrightarrow\qquad f(x)=\left(\frac{3}{5}\right)^x +\left(\frac{4}{5}\right)^x-1=0$$ Clearly, the left hand side is a strictly decreasing function as a sum of two strictly decreasing functions.
Also $$f(2)=0$$, and hence $$x=2$$ is the one and only root of $$f$$.