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Can somebody help me solve this equation?

$x\times 10^{\frac{1}{x}}+\frac1x \times 10^x=20$

What I did: I think the left side is decreasing I tried to show that the first derivative

$10^{\frac1x} - \frac{1}{x^2} 10^x - \frac{1}{x}(10^{\frac1x} - 10^x) \log(10)$ is negative, but I don't know how. How to do it?

amWhy
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1 Answers1

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If $x$ is negative, the left hand side is negative, so no solution.

If $x$ is positive, we can apply AM-GM twice:

$$x\cdot 10^{\frac{1}{x}}+\frac{10^x}{x} \geq 2\sqrt{10^{x+\frac{1}{x}}}\geq 2\sqrt{10^2}=20.$$

The latter inequality is an equality if and only if $x=\tfrac1x$, i.e. iff $x=1$. In this case the former inequality is also an equality, so we get

$$x\cdot 10^{\frac{1}{x}}+\frac{10^x}{x} =20,$$ if and only if $x=1$.

Servaes
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LHF
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  • Can you please demonstrate that equality occurs if and only if $x =1$? I mean, never a good idea to claim a unique solution without proving it, and I see nothing in the way of a proof, other than a claim. – amWhy Feb 26 '20 at 23:05
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    "Equality occurs if " ... But you fail to prove "only if", hence cannot claim "uniqueness" in the solution. I agree that, from your work, $x=1$ is a solution. But you haven't demonstrated that $x=1$ is the unique solution. Maybe a nitpick, but, hey, this is math.se. – amWhy Feb 26 '20 at 23:10
  • @amWhy Maybe my command over the English language is not good. Is the 'only if' part: $x=1 \implies $ the equation or is it: the equation implies $x=1$ ? – LHF Feb 26 '20 at 23:13
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    @Arthur, If you read the proof carefully, I also used $x+\frac{1}{x} \geq 2$. So we also need $x+\frac{1}{x}=2$ to attain the equality. – LHF Feb 26 '20 at 23:15
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    @amWhy, I request you to read the proof one more time. In order to prove that $LHS \ge RHS$ when $x$ is positive, there is an intermediary step $2\sqrt{10^{\frac{1}{x}+x}}\geq 2\sqrt{10^2}$. I'm not sure what's not clear about it. – LHF Feb 26 '20 at 23:22
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    @Arthur and amWhy, I think you guys confuse $10^{\frac{1}{x}}$ with $\frac{1}{10^x}$ and you missed the intermediary usage of AM-GM in the proof. – LHF Feb 26 '20 at 23:24
  • @amWhy, what exactly do I fail to demonstrate ? – LHF Feb 26 '20 at 23:28
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    @amWhy, I edited one more time. I honestly don't know how to explain it more clearly and completely than this. Can you please check again? – LHF Feb 26 '20 at 23:33
  • Okay, @Atticus. I really am not trying to be difficult. So I'll say no more. I appreciate your effort. – amWhy Feb 26 '20 at 23:48
  • @amWhy, please do say more. All I'm asking from you is to pinpoint a specific step like 'It's not clear how the equality case of AM-GM implies $x=1$'. Or, 'I'm not sure how $x\cdot 10^{\frac{1}{x}}+\frac{10^x}{x}\geq 2\sqrt{10^{x+\frac{1}{x}}}$' – LHF Feb 26 '20 at 23:50
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    @Atticus Very nice solution! +1 – Michael Rozenberg Feb 27 '20 at 03:25
  • @AmWhy It is implicitly used that equality holds in AM-GM if and only if all terms are equal. In this case $x=\tfrac1x$ for the latter inequality and $x10^{\tfrac1x}=\tfrac1x10^x$ for the former. – Servaes Feb 27 '20 at 21:45
  • @Atticus I have edited your answer rather liberally in order to clarify that you do indeed prove uniqueness. Feel free to revert or otherwise change my edit to your preference. – Servaes Feb 27 '20 at 21:46
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    @Servaes, It's ok, thanks. I think me and amWhy have pretty much clarified it in the end :) – LHF Feb 27 '20 at 21:48