Question

If $P=\frac{s}{{(1+k)}^n}$, then $n$ equals:

• A. $\frac{\log \left(s/P\right)}{\log (1+k)}$
• B. $\log \frac{s}{P(1+k)}$
• C. $\log \frac{s-P}{1+k}$
• D. $\log \frac{s}{P}+\log (1+k)$
• E. $\frac{\log s}{\log P(1+k)}$

Answer 1

The correct option is A $\frac{\log \left(s/P\right)}{\log (1+k)}$

$(1+k{)}^n=(s/P)$

Taking log on both side,

$n\log (1+k)=\log \left(s/p\right)$

$n=\frac{\log \left(s/p\right)}{\log (1+k)}$