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How to find analytically the third root (-ve) of $x^{2}=2^{x}$ other than 2 and 4?

Does differentiation of the equation make a sensible way?

I tried with $\log_a$ for different $a$' s. But I couldn't find the root.

Riaz
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1 Answers1

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Looking for the negative solution, we get $$ x^2=2^x\\ x^2=e^{x\ln 2}\\ -x=e^{x/2\cdot\ln 2}\\ -xe^{-x/2\cdot\ln 2}=1\\ -\frac x2\ln 2e^{-x/2\cdot\ln 2}=\frac12\ln 2\\ -\frac x2\ln 2=W\left(\frac{\ln 2}2\right)\\ x=-\frac{2W\left(\frac{\ln 2}2\right)}{\ln 2}\approx-0.766665 $$ where the sign on the left-hand side in line 3 comes from knowing that $x$ is negative, and $W$ is the Lambert $W$ function. Which is to say, $W(1/2\cdot\ln 2)$ is the solution to $$ ye^y=\frac{\ln 2}2 $$

Arthur
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    In my opinion you have been here long enough to suspect that this is a duplicate. Did you search? Approach0 gave many hits, granted some in AOPS. – Jyrki Lahtonen Jul 17 '20 at 05:16
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    @JyrkiLahtonen Approach0 still does not work at all on mobile. Neither does the "related questions" column. So I am basically blind in that regard, compared to you. It is, in other words, a lot easier for me to just answer than it is to look for duplicates. Also, you are a lot more concerned about duplicates than I am. I am a lot more concerned about answers in comments than you are. We all have our differences. – Arthur Jul 17 '20 at 05:40
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    While we need to respect differences of opinion, the point raised by @JyrkiLahtonen is important. I also use this site exclusively on mobile and have issues searching, but would prefer to close duplicates than answer them. – Paramanand Singh Jul 17 '20 at 06:24
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    There is also SearchOnMath and that does work on mobile devices. – Martin R Jul 17 '20 at 06:37
  • @MartinR: I have used it on some occasions (you had mentioned it sometime ago to me). While it's not as good as approach0, it is helpful. – Paramanand Singh Jul 17 '20 at 07:29
  • I answer in a comment for many a reason. One of the more frequent ones is that I suspect the question to be a duplicate. – Jyrki Lahtonen Jul 18 '20 at 16:34