Question

Suppose `f(x)= axln(x+b)` where `f(1)=1/2ln3` and `f'(0)=1/2ln2`. Can you find the constants `a` and `b`?

Answer 1

`a=1/2` and `b=2`

Explanation:

Plug `x=1` into `f`:

`f(1)=a(1)ln(1+b)=aln(1+b)=1/2ln(3)`

This would seem to imply that `a=1/2` and `b=2`, but let's check by also finding the derivative through the product rule:

`f'(x)=a(d/dxx)ln(x+b)+ax(d/dxln(x+b))`

`f'(x)=a(1)ln(x+b)+ax(1/(x+b))(1)`

`f'(x)=aln(x+b)+(ax)/(x+b)`

Then:

`f'(0)=aln(b)=1/2ln2`

Which supports our previous theory that `a=1/2` and `b=2`.