# Question-131099

Question Number 131099 by mohammad17 last updated on 01/Feb/21
Commented by mohammad17 last updated on 01/Feb/21
$${how}\:{can}\:{solve}\:{this} \\$$$$\\$$
Answered by mr W last updated on 01/Feb/21
$$\left({a}\right) \\$$$${A}={P}\left(\mathrm{1}+{r}\right)^{{n}} \\$$$$\Rightarrow{n}=\frac{\mathrm{ln}\:\left(\frac{{A}}{{P}}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{r}\right)} \\$$$${n}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{1}.\mathrm{1}}=\mathrm{7}.\mathrm{27}\:{years} \\$$$$\left({b}\right) \\$$$${with}\:{k}\:{compounding}\:{periods}\:{in}\:{one} \\$$$${year}: \\$$$${A}={P}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{{nk}} \\$$$${continuous}\:{means}\:{k}\rightarrow\infty: \\$$$${A}={P}×\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{{nk}} \\$$$${A}={P}×\left[\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{\frac{{k}}{{r}}} \right]^{{nr}} ={Pe}^{{nr}} \\$$$$\Rightarrow{n}=\frac{\mathrm{ln}\:\left(\frac{{A}}{{P}}\right)}{{r}} \\$$$${n}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{0}.\mathrm{1}}=\mathrm{6}.\mathrm{93}\:{years} \\$$
Commented by mohammad17 last updated on 01/Feb/21
$${sir}\:{where}\:{the}\:{row}\:? \\$$
Commented by mohammad17 last updated on 01/Feb/21
$${Professor}\:,{do}\:{you}\:{have}\:{a}\:{page}\:{in}\:{the}\: \\$$$${Facebook}\:{application}\:{or}\:{in}\:{the}\:{telegram} \\$$$${application}\:?\:{i}\:{am}\:\:{communicating}\: \\$$$${with}\:{you}\:.{You}\:{are}\:{really}\:{a}\:{wonderfull} \\$$$${person}\:.\:{can}\:{you}\:{give}\:{me}\:{page}\:{you}\:? \\$$
Commented by mohammad17 last updated on 01/Feb/21
$${can}\:{you}\:{exact}\:{by}\:{details}\: \\$$
Commented by mr W last updated on 01/Feb/21
$${i}\:{have}\:{added}\:{some}\:{explanation}\:{in}\:{the} \\$$$${answer}. \\$$
Commented by mohammad17 last updated on 01/Feb/21
$${thank}\:{you}\:{sir} \\$$