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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?

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I've only skimmed this one but I don't think it has any proof by induction. Still, it could help.… – 2012ssohn Apr 6 '14 at 2:38
Thanks I saw this, however I need to prove this by induction and I don't know how that could be done. – user140751 Apr 6 '14 at 2:45
@user47515 You can't prove this by induction. Not for all $x>0$. You can prove it by induction for $x\in\mathbb N$. – Jack M Apr 6 '14 at 2:57
@user47515 I added the steps for your induction that you want to use. Remember this only proves that $x > \log(x)$ for $x \in \mathbb{N} \setminus \{0\}$. – Rustyn Apr 6 '14 at 3:02
up vote 3 down vote accepted

I don't know why you'd use induction, (unless your domain of each function is $\mathbb{N}\setminus \{0\}$). Here is an alternative approach using calculus. If this is not helpful, I can delete this answer.

Let $g(x)= x- \log(x)$.

$g'(x) = 1 - \frac{1}{x} > 0 $

for all $ x >1$. So $g(x)$ is increasing on $(1,\infty)$.

At $x=1$, $g(x) = 1$, thus $x - \log(x) \ge 0$ for all $x \ge 1$.

Now for $x\in (0,1)$, $\log(x) < 0$ and $x>0$ thus $x-\log(x) > 0$.

Thus $x-\log(x) > 0 $ for all $x \in (0,\infty)$. And conclude $x> \log(x) $ for all $x\in (0,\infty)$.


If you want to use induction to show that for each $x\in \mathbb{N}\setminus \{0\}$, $x>\log(x)$, use your inductive hypothesis via: $$ k > \log(k) \longrightarrow \\ k+\log(1+\frac{1}{k})> \log(k)+\log(1+\frac{1}{k}) = \log(k+1) \\ k+\log(1+\frac{1}{k}) \le k + \log(2) \text{ and } \log(2) < 1 \text{ so } \\ k + \log(2) < k + 1 \text{ thus } \\ k+1 > k + \log(2) \ge k + \log(1+\frac{1}{k}) > \log(k+1) $$ Q.E.D.

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The problem I'm trying to solve explicitly says to use induction. The calculus approach makes more sense to me though. Also don't delete your answer it could be helpful to others! – user140751 Apr 6 '14 at 2:44
@user140751 But the functions $x$, $\log x$ are usually defined over subsets of $\mathbb{R}$. Induction really only makes sense if you are talking about the natural numbers. If you prove $x > \log(x)$ for $x$ a natural number,you still haven't proved $x>\log(x)$ where $x$ is a real between two natural numbers. – Rustyn Apr 6 '14 at 2:46

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 $\log{2}<1$

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$\log(ab) \neq \log(a)\cdot \log(b)$. Consider $a = e$ and $b = 1$. Then $\log(ab) = \log(e) = 1$, but $\log(a)\cdot \log(b) = 1\cdot 0 = 0$. You are mistaking the correct identity: $\log(ab) = \log(a)\cdot \log(b)$. – Chris K Apr 6 '14 at 2:50
@ChrisK I think you mean $\log(ab) = \log(a) + \log(b)$ for the "correct identity." ;) – apnorton Apr 6 '14 at 2:53
Thank you! Brain fart. I wrote down the mistake. Yes, indeed. – Chris K Apr 6 '14 at 2:55
That would be the case! I did finish the inductive proof now. – Stella Biderman Apr 6 '14 at 2:56

(I) Let $x\in \mathbb R, x>0$. Maclaurin series for $e^x$: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$,
So $e^x-x=1+\frac{x^2}{2!}+\frac{x^3}{3!}+...>0$ or $e^x>x$ or $x>log(x)$. When $x\in \mathbb N^+\subset\mathbb R$ , the statement is still true. This completes the proof.

(II) If $x\in \mathbb R$
Let $g(x)=x-log(x)$.
Since $g(1)=1$,and $g'(x)=1-\frac1x>0$ when $x>1$,
on the interval $(1, \infty)$, $g(x)>1>0$,
or $x>log(x)$.
On the interval $(0,1), x>0, log(x)<0$, so $x>log(x)$,
when $x=1$, then $log(x)=0<1=x$, so $x>log(x)$. Since $x\in \mathbb N^+ \subset \mathbb R$ , The statement is still true for $x>0, x\in \mathbb N$.
This completes the proof.

(III)If x $\in \mathbb R$
Since $e^t|_{t=0}=1$, and $e^t$ is monotonically increasing in interval $(0,\infty)$, Thus $e^t>1$ is always true when t>0.
so $\int^x_0e^tdt> \int^x_0dt$, for $x>0$ or $1+\int^x_0e^tdt> \int^x_0dt$ or $e^x>x$ when $x>0$
or $e^x>x>0$,
or $1>\frac x{e^x}>0$
or $log (\frac x{e^x})<0$,
or $log(x)-log(e^x)<0$
or $log(x)<log(e^x)$ or $log(x)<x$ when $x>0$.
Since $x\in \mathbb N^+ \subset \mathbb R$ , The statement is still true. This completes the proof.

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