Prove that $e^x - \ln(x)=2^{2014}$ has exactly two positive real roots.
I know how to do it by graph. I need equation based solution.
I differentiated it twice and got it's strictly increasing on $0$. How to proceed?
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User12345
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You have two tools: 1) the intermediate value theorem can be used to show that a function has at least one solution on a given interval, and 2) rolle's theorem can be used to show that a strictly increasing/decreasing function has at most one zero on a given interval. Put these two things together by finding the right intervals to apply 1) and 2) to the function $f(x) = e^x - \ln(x)-2^{2014}$. – AnonymousCoward Dec 27 '17 at 12:45
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1There is also a same question in diff. post herethis post – BAYMAX Dec 27 '17 at 12:53
2 Answers
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Because $f(x)=e^x-\ln{x}$ is a convex function,
$f(1)<2014$, $\lim\limits_{x\rightarrow0^+}f(x)=+\infty$ and $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$.
Thus, there are $0<a<1$ and $b>1$ for which $f(a)=f(b)=2014$, but since $f$ is a convex function, the equation $f(x)=2014$ has no another roots.
Michael Rozenberg
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I need to prove it by equations, find limit and show it's answer is from between [a,b] & [c,d] – User12345 Dec 27 '17 at 12:40
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I showed that in $(0,1)$ there is an unique root and in $(1,+\infty)$ there is an unique root. – Michael Rozenberg Dec 27 '17 at 12:43
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Actually how we get $\lim_{x \rightarrow 0^{+}} f(x) = + \infty$ ? like its of the form $1 - \ln(0) $ like $-\infty$ ? @MichaelRozenberg ? – BAYMAX Dec 27 '17 at 12:52
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@BAYMAX Because $\lim\limits_{x\rightarrow0^+}(-\ln{x})=+\infty$. – Michael Rozenberg Dec 27 '17 at 12:55
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I understood this part -> "Thus, there are $0<a<1$ and $b>1$ for which $f(a)=f(b)=2014$" but after this why convexity plays role? what would have happened if it was not convex? – BAYMAX Dec 27 '17 at 13:26
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@BAYMAX By definition of the convex function $f(x)<2014$ for all $a<x<b$ and $f(x)>2014$ for all $x\in(0,1)\cup(b,+\infty)$, which says that only $a$ and $b$ they are roots. – Michael Rozenberg Dec 27 '17 at 14:17
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If we put $f(x)=e^x-\ln(x)-2^{2014}$, we need to show how many times it cut f(x)=0. At first, his minimal is at $f'(x)=e^x-1/x=0$, and this is happen at $x=W(1) \approx 0.567143$. We can see then that $f(W(1))<0$ Because $f''(x)>0$ for all $x>0$, and $\lim_{x \to 0+} f(x)=\infty$, and $\lim_{x \to \infty} = \infty$, then it can have only 2 places which pass in $f(x)=0$, one is at $0<x<W(1)=0.567143$ and other is at $x>W(1) \approx 0.567143$
If first root a is at (0,0.567143). Second root b is at $(ln(2^{2014}),ln(2^{2015}))=(1395.998421648,1396.691568828)$ because $f(\ln(2^{2014}))=e^{\ln(2^{2014})}-\ln(\ln(2^{2014}))-2^{2014}=2^{2014}-\ln(\ln(2^{2014}))-2^{2014}<0$
Daniel Pol
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