I am trying to find the solution of this equation:
$$2x=5^{x-1}$$
I have rearranged it but not made any progress.
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2$\begingroup$ Please include in the question what exactly you tried and why you are interested in the solution (using the Edit button under the question). Otherwise you will receive poor feedback and the question may be closed. $\endgroup$– MatijaAug 31 at 20:07
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$\begingroup$ You should also write about the work that you have done so far on this problem. Every post should ideally contain accompanying work. $\endgroup$– ShyamalSayakAug 31 at 23:16
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1 Answer
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Let $W_n(x)$ the lambert function (the inverse function of $x e^x$)
$$2x=5^{x-1}$$
$$10x=5^x$$
$$10x=e^{\ln(5) x}$$
$$10 x e^{-\ln(5)x}=1$$
$$-10 \ln(5)x e^{-\ln(5)x}=-\ln(5)$$
$$-\ln(5)x e^{-\ln(5)x}=-\frac{\ln(5)}{10}$$
$$-\ln(5)x=W_n\left(-\frac{\ln(5)}{10}\right)$$
$$x=-\frac{W_n\left(-\frac{\ln(5)}{10}\right)}{\ln(5)}$$
You have real solutions for $n=-1$ and $n=0$
$$x\approx 0.121621\text{ and }x\approx 1.79372$$
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1$\begingroup$ @user $xe^x$ has more than one possible inverse, based on the $n$ you choose you are deciding a branch. en.wikipedia.org/wiki/Lambert_W_function $\endgroup$ Aug 31 at 20:11
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$\begingroup$ @MathAttack First sight, question was looking like high school level. I find your answer very interesting do you think there's another way to solve the equation or It's one kind that looks harmless and turn out to be an hydra? I'm asking out of curiosity $\endgroup$ Aug 31 at 21:18
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1$\begingroup$ @TurquoiseTilt surely it can be solved numerically with some recursive method, but you still arrive at an approximate solution and not an exact solution. $\endgroup$ Aug 31 at 23:35
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2$\begingroup$ @TurquoiseTilt Personally I don't find it a trivial question, the Lambert function is not studied in high school. I saw it for the first time at the university, the same thing goes for the iterative methods to numerically solve this kind of equations. I don't understand why they are downvoting the question (apart from lack of context perhaps), the question is simple but the answer is not obvious. $\endgroup$ Aug 31 at 23:38