Question

$\frac{a-b}{a}+\frac{1}{2}{\left(\frac{a-b}{a}\right)}^2+\frac{1}{3}{\left(\frac{a-b}{a}\right)}^3+...$ to $\infty$ is equal to

• A. $\log a+\log b$
• B. $\log \frac{b}{a}$
• C. $\log a-\log b$
• D. none of these

Answer 1

The correct option is C $\log a-\log b$

$\log \left(x\right)=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}...\infty$
Simplifying the above expansion, we get
$(1-\frac{b}{a})+\frac{1}{2}(1-\frac{b}{a}{)}^2+...\infty$
$=-\log (1-(1-\frac{b}{a}\left)\right)$
$=-\log \left(\frac{b}{a}\right)$
$=\log \left(\frac{a}{b}\right)$
$=\log \left(a\right)-\log \left(b\right)$