Problem: If P=s(1+k)nP=\dfrac{s}{(1+k)^{n}}P=(1+k)ns, then nnn equals:
Answer Choices:
A. logs/Plog(1+k)\dfrac{\log s / P}{\log (1+k)}log(1+k)logs/P
B. logsP(1+k)\log \dfrac{\mathrm{s}}{\mathrm{P}(1+\mathrm{k})}logP(1+k)s
C. logs−P1+k\log \dfrac{s-P}{1+k}log1+ks−P
D. logsP+log(1+k)\log \dfrac{s}{P}+\log (1+k)logPs+log(1+k)
E. logslogP(1+k)\dfrac{\log s}{\log P(1+k)}logP(1+k)logs Solution:
logP=logs−nlog(1+k)\log P=\log s-n \log (1+k) logP=logs−nlog(1+k)
n=logs−logPlog(1+k)=logs/Plog(1+k)n=\dfrac{\log s-\log P}{\log (1+k)}=\dfrac{\log s / P}{\log (1+k)} n=log(1+k)logs−logP=log(1+k)logs/P