High School Why does the expression equal the reciprocal of its logarithm?

[terryds]

I encountered this in http://calcchat.com/book/Calculus-10e/8/4/7/

How come the above expression equals the below?
What I know it should be 4 ln(x/(4+sqrt(16-x^2))) which means the -1 becomes the power of that thing inside ln.

Please help me. I really don't get it.

 

[andrewkirk]

Provided $x\in[-4,0)\cup (0,4]$ we have
$$\left|\frac{4+\sqrt{16-x^2}}x\right|=\frac{4+\sqrt{16-x^2}}{|x|}$$
and
$$\left|\frac{4-\sqrt{16-x^2}}x\right|=\frac{4-\sqrt{16-x^2}}{|x|}$$
and that multiplying the two right-hand sides together gives 1. So they are reciprocals, hence their logs are additive inverses.