Question

What is the value of `cosh(ln3)`?

Answer 1

`cosh(ln3)=5/3`

Explanation:

`Coshx=1/2(e^x+e^-x) AAx inRR`

so

`cosh(ln3)=1/2(e^(ln3)+e^-ln3)`

`cosh(ln3)=1/2(e^(ln3)+e^ln(1/3))`

`cosh(ln3)=1/2(3+1/3)=1/2(10/3)`

`cosh(ln3)=5/3`