# Math Help - Natural log problem

1. ## Natural log problem

Show that x2 - 12x + 11 + 18ln(x) > 0, for x > 1.

My attempt:

I start by evaluating x2 - 12x + 11 = p.
The expression equals zero at x = 1, x = 11. It is positive for all x > 11 and negative on 11 > x > 1.

Since ln(x) > 0 for every x > 1, and the expression p is positive when x > 11; the inequality holds for x > 11.

I'm guessing the next step is to show that (x-11)(x-1) > -18ln(x) for 11 > x > 1.

But I don't know how to proceed from here

2. ## Re: Natural log problem

Originally Posted by Cinnaman
Show that x2 - 12x + 11 + 18ln(x) > 0, for x > 1.
I suggest that you consider $f(x)=x^2-12x+11+18\ln(x)$.

Note that $f(1)=0$ and if $x>1$ then $f'(x)>0$.