In an attempt to find the solution to the equation
$Mx=e^x$ with $M$ being a large real number and the solution $w \gt 1$
I was asked to justify why $\ln M$ is a reasonable approximation to $w$.
I was also asked to improve the approximation by writing $w=\ln M +y$. I need to find an approximation for $y$ hence improve the approximation $\ln M$.
I vaguely know that $\ln M$ is reasonable because $M$ is large, but feel a bit uncertain, and I do not quiet understand the second part of the question. Could somebody help me please?
$\ln M$ is a rough approximation because it turns the dominant term, $e^x$, into $M$; so we are off by a factor of $\ln M$ "only": $M\ln M\ne e^{\ln M}$.
What could be a good choice of $y$ to make $w=\ln M+y$ a better appoximation? Just as the first approximation dealt with the factor $M$ on the left, we may want choose $y$ so that it deals with the (now) secdond factor $\ln M$; that is, we let $y=\ln\ln M$.This we we find $M\ln w=M\ln M+M\ln\ln M$ and $e^w=e^{\ln M+\ln\ln M}=M\ln M$.
Another way to put this and at the same time to see how to continue to a full sequence of improving approximations: Apply $\ln$ to both sids of the original equation to get $$x=f(x)=\ln M+\ln x. $$ Now we are looking for a fixed point of $f$ and - at least in the range we are interested in - we have $|f'(x)|=\frac1x\approx \frac1{\ln M}\ll 1$. Therefore the sequence defined by $x_1=\ln M$, $x_{n+1}=f(x_n)$ converges quite nicely to a limit $w$ that solves the original equation. Actually, it doesn't even matter a lot what we pick as first approximation $x_1$ (for example, we get the very same sequence with just one additional term if we start with the very different starting point $x_0=1$).