 David Young

2022-01-21

Show ${\left(\mathrm{log}n\right)}^{\left(\mathrm{log}n\right)}={2}^{\left(\mathrm{log}n\right)\left(\mathrm{log}\left(\mathrm{log}n\right)\right)}$
I am having difficulty understanding how this follows.
${\left(\mathrm{log}n\right)}^{\left(\mathrm{log}n\right)}={2}^{\left(\mathrm{log}n\right)\left(\mathrm{log}\left(\mathrm{log}n\right)\right)}={n}^{\mathrm{log}\mathrm{log}n}$
Which logarithmic identities are used to go through each equality?
e.g. how do you first go from
${\left(\mathrm{log}n\right)}^{\left(\mathrm{log}n\right)}={2}^{\left(\mathrm{log}n\right)\left(\mathrm{log}\left(\mathrm{log}n\right)\right)}$
and then to
${2}^{\left(\mathrm{log}n\right)\left(\mathrm{log}\left(\mathrm{log}n\right)\right)}={n}^{\mathrm{log}\mathrm{log}n}$
(The log base must be 2 or else this equality won't hold) enhebrevz

Expert

Im Barbara Meeker

Expert

For any
$1\ne a\in {\mathbb{R}}^{+},{x}^{y}={a}^{y{\mathrm{log}}_{a}x}$,whenever LHS is defined.
Thus,
${\left(\mathrm{log}n\right)}^{\mathrm{log}n}={2}^{\mathrm{log}n\mathrm{log}2\left(\mathrm{log}n\right)}$
So your equality follows if the logarithm here is taken in base $2$ and not $10$ , as you wrote... RizerMix

Expert

Let $y={\mathrm{log}}_{2}^{{\mathrm{log}}_{2}n}n$ Taking the logarithm (in base 2) of both sides ${\mathrm{log}}_{2}y={\mathrm{log}}_{2}n{\mathrm{log}}_{2}{\mathrm{log}}_{2}n$ Now, remember that ${2}^{{\mathrm{log}}_{2}n}=n$. Thus $y={2}^{{\mathrm{log}}_{2}y}={2}^{{\mathrm{log}}_{2}n{\mathrm{log}}_{2}{\mathrm{log}}_{2}}$ Also recall that ${a}^{bc}=\left({a}^{b}{\right)}^{c}$. Thus $y={2}^{{\mathrm{log}}_{2}n{\mathrm{log}}_{2}{\mathrm{log}}_{2}n}=\left({2}^{{\mathrm{log}}_{2}n}{\right)}^{{\mathrm{log}}_{2}{\mathrm{log}}_{2}n}={n}^{{\mathrm{log}}_{2}{\mathrm{log}}_{2}n}$