Solve for $n$ in $2^Mn=1.00001^n$

Question

If $2^Mn=1.00001^n$, how to compute $n$ in respect of $M$? Assume $M$ is a constant.

I don't know how to represent $n$ by $M$. Any suggestions? Might need to apply logarithms or root somehow.

Answer 1

The equation is of the form $kx=a^x$ with $k,a$ constant. This cannot be solved for $x$ with logarithms or roots alone. Rather, a very powerful special function – the Lambert W (product logarithm) – is needed, defined as follows: $$x=W(x)e^{W(x)}$$ Many equations involving linear and exponential terms can be solved this way, and the Wikipedia page I linked conveniently has an example of how to solve $kx=a^x$: $$kx=a^x=e^{x\ln a}$$ $$kxe^{-x\ln a}=1$$ $$-x\ln a\cdot e^{-x\ln a}=-\frac{\ln a}k$$ $$-x\ln a=W\left(-\frac{\ln a}k\right)$$ $$x=-\frac{W\left(-\frac{\ln a}k\right)}{\ln a}$$ Substituting $n=x$, $a=1.00001$ and $k=2^M$ leads to the solution for the given equation: $$n=-\frac{W\left(-\frac{\ln1.00001}{2^M}\right)}{\ln1.00001}$$ The top branch of the Lambert W yields a solution approximately equivalent to $2^{-M}$ for $M\gtrsim-5$. The bottom branch, as mentioned in the comments of this answer, skyrockets for small values of $M$.