Suppose $f (x) = a x ln (x + b)$ $f(x)= axln(x+b)$ where $f (1) = \frac{1}{2} ln 3$ $f(1)=1/2ln3$ and $f'(0)=1/2ln2$ . Can you find the constants $a$ and $b$ ?

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Answer 1 · 2017-02-17T12:10:18

$a=1/2$ and $b=2$

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 $b=2$ .

Suppose f(x)= axln(x+b)f(x)=axln(x+b)f(x)= axln(x+b) where f(1)=1/2ln3f(1)=12ln3f(1)=1/2ln3 and f'(0)=1/2ln2f'(0)=1/2ln2. Can you find the constants aa and bb?