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This question already has an answer here:

Show that the equation $$3^x+4^x=5^x$$ has exactly one root.

Consider x as a member of real number set. This question was given by my instructor as part of Calculus course. I don't have any clue on how to approach this. Please help me in solving this.

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marked as duplicate by lab bhattacharjee calculus Oct 21 '17 at 0:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Here’s a hint: rearrange so that the equation has 0 on one side. Now differentiate. If the derivative is always positive or always negative then the function must have at most one root so find a value where it’s positive and where it’s negative and you’re done by MVT $\endgroup$ – Dan Robertson Oct 20 '17 at 21:54
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The hint:

We need to solve that $$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1$$

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    $\begingroup$ Further hint: the LHS is a decreasing function of the $x$ variable. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 21:55
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Roots of the given equation correspond to solutions of the equation $$(\frac{3}{5})^x+(\frac{4}{5})^x=1$$ However, the function $f(x)=(\frac{3}{5})^x+(\frac{4}{5})^x$ is decreasing for all $x$ (as the sum of two decreasing functions), hence there is at most one solution. Now find one solution by inspection ($x=2$) and we're done.

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