Question

What is `cosh(ln(x))`?

Answer 1

`cosh(ln(x)) = (x^2+1)/(2x)`

Explanation:

`cosh(z) = (e^z+e^-z)/2`

So:

`cosh(ln(x)) = (e^(ln(x))+e^(-ln(x)))/2 = (x+1/x)/2 = (x^2+1)/(2x)`