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Find $x\in\mathbb{R}$ such that

$$5^{2x-1}-6e^{x-1}+1=0$$

I think that is not possible to find an answer, but I'm not sure.

yemino
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    Well, there is an obvious root coming from the observation that $5+1=6$. – dxiv Jun 26 '18 at 02:59
  • and, is it possible to calculate the other solution? (the plot shows a negative solution too) – yemino Jun 26 '18 at 03:19
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    I don't think the other solution has an elementary description, although it might be possible using the Lambert W function. Numerically it is approximately $-0.775410212890783968$. – Mario Carneiro Jun 26 '18 at 03:22
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    @yemino You can prove that there is one other (negative) solution, and you can approximate it numerically as accurately as needed, but I don't think there is a closed form for it. – dxiv Jun 26 '18 at 03:24
  • http://www.wolframalpha.com/input/?i=solve+5%5E(2x-1)-6*exp(x-1)%3D-1 – saulspatz Jun 26 '18 at 03:48
  • @dxiv you mean $x=1$? – Mr Pie Jun 26 '18 at 04:44
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    @user477343 Yes, $,5^1 - 6 \cdot e^0 + 1 = 0,$. – dxiv Jun 26 '18 at 04:50

1 Answers1

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Consider that you are looking for the zero(s) of function $$f(x)=5^{2x-1}-6e^{x-1}+1$$ for which $$f'(x)=2\ 5^{2 x-1} \log (5)-6 e^{x-1}$$ $$f''(x)=4\ 5^{2 x-1} \log ^2(5)-6 e^{x-1}$$ If you notice that the first derivative can write before simplifications as $$f'(x)=6e^{x-1}\left(\exp\left[(2x-1)\log(5)+\log(2)+\log(\log(5))-(x-1)-\log(6) \right]-1 \right)$$ it cancels at $x_*$ (which makes the argument of the exponential term equal to $0$) given by $$x_*=\frac{\log \left(\frac{15}{ \log (5)}\right)-1}{2 \log (5)-1}\approx 0.555311$$ At this point $$f(x_*)\approx-1.65127 \qquad \text{and} \qquad f''(x_*)\approx 8.53411$$ So, two roots and one of them is obvious.

For the second one, you need a numerical method such as Newton; but it requires a reasonable starting point. By inspection or graphing, you will notice that the second solution is between $0$ and $-1$ since $f(0)=\frac{6}{5}-\frac{6}{e}\approx -1.00728 $ and $f(-1)=\frac{126}{125}-\frac{6}{e^2}\approx 0.195988$.

So start the iterations using $x_0=-1$ and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -1.0000000000 \\ 1 & -0.7507336913 \\ 2 & -0.7751463191 \\ 3 & -0.7754101823 \\ 4 & -0.7754102129 \end{array} \right)$$ which is the solution for ten significant figures.

Claude Leibovici
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