How do I solve this for $x$? [closed]

I am trying to find the solution of this equation:

$$2x=5^{x-1}$$

I have rearranged it but not made any progress.

• 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. Aug 31 at 20:07
• You should also write about the work that you have done so far on this problem. Every post should ideally contain accompanying work. Aug 31 at 23:16

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 $$n=0$$
$$x\approx 0.121621\text{ and }x\approx 1.79372$$
• What is the meaning of $n$?
• @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 Aug 31 at 20:11