I'm trying to prove $ \log(x) < x$ for $x > 0$ by induction.
Base case: $x = 1$
$\log (1) < 1$ —> $0 < 1$ which is certainly true.
Inductive hypothesis: Assume $x = k$ —> $\log(k) < k$ for $k > 0$
Inductive conclusion: Prove $\log(k+1) < k+1$
I don't know what to do after this. I mean the statement itself is quite obviously true, but how do I continue with the proof?
Best Answer
Induction only works for integers. The easiest way to prove this is to note that $e^x>x$ (The power series for $e^x$ is only positive terms and one of them is $x$), and then let $x=\ln{y}$.
For a proof by induction, factoring $k$ out, yields $\ln{(k+1)}=\ln{k}+\ln{(1+\frac{1}{k})}<k+\ln{(1+\frac{1}{k})}<k+1$ since $\log{2}<1$