How can I show that the equation $$e^x-\ln(x)-2^{2014}=0$$ has exactly two positive roots?
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$\ln(\alpha) < 0 \quad \forall \alpha\in(0,1)$... So your thoughts are wrong – AlexR Jan 05 '14 at 18:47
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It was a good edit!! – Mikasa Jan 05 '14 at 18:49
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Define $f(x) = e^x - \ln x$. Then
$$f'(x) = e^x - \frac 1 x \implies f''(x) = e^x + \frac 1 {x^2} \ge 1$$
for all $x > 0$. That is, $f$ is concave up, so it's shaped like the letter $U$. Hence there are at most two positive roots. Now use the intermediate value theorem.
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It goes to infinity at both $0$ and $\infty,$ and the derivative is $\exp x - 1/x,$ which has exactly one zero, and the function is negative at that zero.
Igor Rivin
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