# Ask Uncle Colin: a proof with logs in

*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions – and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I have a slightly embarrassing problem: I can’t seem to show that $\ln(x) < x$ for all $x>0$, no matter how hard I try. Can you explain?

-- Not All Proofs Involve Exponential Ratios

Dear NAPIER,

Of course I can explain! I like this question. It’s one of those “well… obviously!” questions that a look at a graph would show beyond any doubt, but *that doesn’t count as a proof*. I have a few approaches that work, though.

### Dealing with $0\lt x\le1$

It’s trivially true when $x$ is no bigger than 1: in that domain, $x$ is positive and $\ln(x)$ isn’t, so there’s no contest, $x > \ln(x)$ there.

### The spot-the-difference approach

This is probably the simpler of the two to work out. An alternative to showing $\ln(x) < x$ is to show that $x - \ln(x) > 0$. That’s clearly true when $x \le 1$, but what about for larger $x$? If we differentiate, we get $1 - \frac 1x$, which is positive for $x>1$ – so $x - \ln(x) = 1$ when $x=1$ and then increases in value; therefore, $x>\ln(x)$ for all $x$.

### The Maclaurin series approach

There’s another, more complicated approach that I like. To show $x > \ln(x)$, you can show $e^x > x$. The Maclaurin series expansion of $e^x$ is $e^x = 1 + x + \frac 1{2!} x^2 + \frac 1{3!}x^3 + …$, which is made of positive terms, so it’s clearly greater than $x$.

### The infinite regress

A proof by contradiction: suppose there’s a real solution, $a$ to $x = \ln(x)$. Clearly, $x > 1$ from before.

However, since $a = \ln(a)$, $a = \ln(\ln(a))$. The right hand side is only positive for $x > e$, so $a > e$

Similarly, you can argue that $a = \ln(\ln(\ln(a)))$ – and the right hand side is only positive if $x > e^e$, so $a > e^e$.

You can carry on like this forever – and show that $a$ has to be bigger than any real value you might care to pick, meaning there’s no such $a$.

-- Uncle Colin

* Edited 2015-08-05 to fix some LaTeX.