please could you help with this question.
If a and b are small compared with x, show that
$$ln(x+a) - lnx = \frac{a}{b}(1 + \frac{b-a}{2x})(ln(x+b) - lnx)$$
I've tried expanding ln(x+a) as a taylor series but am unsure how to proceed from there.
$$ln(x+a) - lnx = \frac{a}{x} - \frac{a^2}{2x^2} + \frac{a^3}{3x^3} - \frac{a^4}{4x^4} +...$$
Thanks.
Hint
Just as you did, consider the lhs $$\log(x+a) - \log(x) = \log\frac{x+a}{x}=\log(1+\frac{a}{x})$$ Use now the fact that if $y$ is small $$\log(1+y) \approx y-\frac{y^2}{2}+\frac{y^3}{3}+\cdots$$ and replace $y$ by $\frac{a}{x}$. So,$$\log(x+a) - \log(x) \approx \frac{a}{x}-\frac{a^2}{2x^2}+\frac{a^3}{3x^3}$$ Just do the same for $$\log(x+b) - \log(x)\approx\frac{b}{x}-\frac{b^2}{2x^2}+\frac{b^3}{3x^3}$$ So $$\frac{\log(x+a) - \log(x) }{\log(x+b) - \log(x) }\approx\frac{\frac{a}{x}-\frac{a^2}{2x^2}+\frac{a^3}{3x^3}}{\frac{b}{x}-\frac{b^2}{2x^2}+\frac{b^3}{3x^3} }=\frac{a-\frac{a^2}{2x}+\frac{a^3}{3x^2}}{b-\frac{b^2}{2x}+\frac{b^3}{3x^2} }$$ Now perform the long division.
I am sure that you can take from here.