# Prove the inequality for all natural numbers n using induction log_2 n<n I know how to prove the base case Base Case log_2 1<1 likewise assuming the inequality for n=k; log_2 k<k Then to prove by induction I show log_2 k<(k+1)?

Prove the inequality for all natural numbers n using induction
${\mathrm{log}}_{2}n
I know how to prove the base case Base Case ${\mathrm{log}}_{2}1<1$ likewise assuming the inequality for n=k; ${\mathrm{log}}_{2}k
Then to prove by induction I show ${\mathrm{log}}_{2}k<\left(k+1\right)$?
I know it's true since the domain is all real numbers i just cant figure out the next step to prove it.
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Liam Everett
${\mathrm{log}}_{2}\left(n\right)=\frac{\mathrm{ln}\left(n\right)}{\mathrm{ln}\left(2\right)}$, where $ln$ is the natural log i.e. with base $e$
You need to show, ${\mathrm{log}}_{2}\left(n\right)
Since $\mathrm{log}$ is an increasing function you can rephrase your question to $n<{2}^{n}$
That is easy to show by induction
n=1: $1<{2}^{1}$. True.
Induction Hypothesis: $n-1<{2}^{n-1}$
for $n\ge 2$, $n\le 2\left(n-1\right)<2\ast {2}^{n-1}={2}^{n}$
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tikaj1x
For $k>1$, we have
$k+1<2k,$
hence, because $\mathrm{log}\left(x\right)$ is increasing
${\mathrm{log}}_{2}\left(k+1\right)<{\mathrm{log}}_{2}\left(2k\right)={\mathrm{log}}_{2}\left(k\right)+1