# Prove (log n)^2<=2^n by induction

Prove $\left(\mathrm{log}n{\right)}^{2}\le {2}^{n}$ by induction
I've trying to solve this for quite a while now, but not being able to finish the proof.
Prove using induction that $\left(\mathrm{log}n{\right)}^{2}\le {2}^{n}$
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Lilliana Mason
For $n=1,2,3$ it's clearly true.
Assume that it holds for $n=k$: $\mathrm{log}k\le {2}^{n/2}$
Then
$\mathrm{log}\left(k+1\right)=\mathrm{log}k\cdot \frac{\mathrm{log}\left(k+1\right)}{\mathrm{log}k}=\mathrm{log}k\left(1+\frac{\mathrm{log}\left(1+\frac{1}{k}\right)}{\mathrm{log}k}\right)\le \mathrm{log}k\left(1+\frac{1}{k\mathrm{log}k}\right)\le {2}^{n/2}\cdot {2}^{1/2}={2}^{\left(n+1\right)/2},$
as
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videosfapaturqz
$\mathrm{log}n for all $n$
${n}^{2}\le {2}^{n}$ for $n\ne 3$
handle the case $n=3$ separately.