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How do I prove that for every positive real number $a$, the equation $^{-}=a$ has a real solution $x>0$?

What I tried: I tried moving everything to one side leaving the other side as $0$. Then I drew up a graph of the $e^{-x}$ and the $ax$ line (where a is some positive gradient) as separate curves. My teacher suggested I substitute $f(0)$ and $f(1/a)$ into the equation to see the sign of the function $f(x)= ax- ^{-x}$. But I don't know how that helps. In general, does anyone have tips for proving this sort of thing, and what category of proof is this?

Vasili
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thetrueembodimentofstupidity
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1 Answers1

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$f(0) = -1 < 0$, $f(1/a) = 1 - e^{1/a} > 0$, and $f$ is continuous. Thus, $f$ is continuous function that has values of different signs on $[0, 1/a]$, and so, by Bolzano's theorem, it has a zero in this interval.

In general, I would start by checking values at ends: in equation $e^{-x} = ax$, if $x = 0$ then left side is bigger (as it's positive and right is zero), but if $x$ is very large, then right side is bigger (as it grows, while left side goes to $0$). As both sides are continuous, there will be an intersection.

(the second paragraph is essentially the same as the first, just without finding explicit value where right side is bigger)

mihaild
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