Question

If $c\left(a-b\right)=a\left(b-c\right)$ then $\frac{\log \left(a+c\right)-\log \left(a-2b+c\right)}{\log \left(a-c\right)}$ equals
(Assume all terms are defined)

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given $c(a-b)=a(b-c)$

So, $(a-b)=\frac{a}{c}(b-c)$(equation $1$)

$\because \frac{\log (a+c)-log(a-2b+c)}{\log (a-c)}$

From Equation $1$

$=\frac{\log (a+c)-log(a+c-\frac{2\cdot 2ab}{(a+c)})}{\log (a-c)}$

$=\frac{\log (a+c)+log\left(\right(a+c{)}^2-4ab)-\log (a+c)}{\log (a-c)}$

$=\frac{\log (a-c{)}^2}{\log (a-c)}=2$