Logarithms- Doubt

Question

I came across this question recently:

I'm completely lost: All I can arrive at is that x = ln(M) +ln(x).

How do I solve this?

Answer 1

If $M$ is large, $\log M$ will not be too small and $x$ is a little larger than $\log M$. $\log x$ will be rather small. You can use fixed point iteration by setting $x_0=\log M, x_{i+1}=\log M +\log x_i$ which will converge quickly. Try it in a spreadsheet. The first iteration is what they are looking for in $y$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{Mx = \expo{x}\quad\imp\quad x = \ln\pars{x} + \ln\pars{M}}$.

1. Let's assume you have two roots $\ds{1 < w_{1} < w_{2}}$. By MVT, you'll get $\ds{1 = {\ln\pars{w_{1}} - \ln\pars{w_{2}} \over w_{1} - w_{2}} = {1 \over \xi}}$ where $\ds{w_{1} < {1 \over \xi} < w_{2}}$. The contradiction $\ds{\pars{~w_{1} < 1~}}$ proves that you have just one solution for $\ds{x > 1}$.
2. $\ds{\ln\pars{M}}$ is a 'good approximation' whenever $\ds{M\ln\pars{M}} \approx \expo{\ln\pars{M}}$ which occurs whenever $\ds{M \approx \expo{}}$. By expanding $\ds{Mx - \expo{x}}$ 'around' $\ds{x = \ln\pars{M}}$: \begin{align} 0 & \approx M\ln\pars{M} - \expo{\ln\pars{M}} + \bracks{M - \expo{\ln\pars{M}}}\bracks{x - \ln\pars{M}} + \half\pars{-\expo{\ln\pars{M}}} \bracks{x - \ln\pars{M}}^{2} \\[5mm] 0 & \approx M\ln\pars{M} - M - \half\,M\bracks{x - \ln\pars{M}}^{2} \quad\imp\quad x \approx \ln\pars{M} \pm \root{2\bracks{\ln\pars{M} - 1}} \end{align}

   ---

   Since $\ds{x > 1}$, the root satisfies $$ x \approx \ln\pars{M} +\ \underbrace{\root{2\bracks{\ln\pars{M} - 1}}}_{y}\,,\qquad M \gtrsim \expo{} $$