Question

If $x=\frac{y}{(1+p{)}^q}$, then $q$ is equal to

• A. $\log \left[\frac{\frac{y}{x}}{1+p}\right]$
• B. $\frac{\log \left(\frac{y}{x}\right)}{\log (1+p)}$
• C. $\frac{\log (y-x)}{\log (1+p)}$
• D. $\frac{\log (y+x)}{\log (1+p)}$

Answer 1

The correct option is B $\frac{\log \left(\frac{y}{x}\right)}{\log (1+p)}$

Given that $x=\frac{y}{(1+p{)}^q}$

Taking logarithm for both the sides, we have

$\log x=\log \frac{y}{(1+p{)}^q}$

$\Rightarrow \log x=\log y-\log (1+p{)}^q$

$\Rightarrow \log x=\log y-q\log (1+p)$

$\Rightarrow q\log (1+p)=\log y-\log x$

$\Rightarrow q=\frac{\log y/x}{\log (1+p)}$