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
![(a) A=P(1+r)^n ⇒n=((ln ((A/P)))/(ln (1+r))) n=((ln 2)/(ln 1.1))=7.27 years (b) with k compounding periods in one year: A=P(1+(r/k))^(nk) continuous means k→∞: A=P×lim_(k→∞) (1+(r/k))^(nk) A=P×[lim_(k→∞) (1+(r/k))^(k/r) ]^(nr) =Pe^(nr) ⇒n=((ln ((A/P)))/r) n=((ln 2)/(0.1))=6.93 years](Q131106.png)
$$\left({a}\right) \\ $$$${A}={P}\left(\mathrm{1}+{r}\right)^{{n}} \\ $$$$\Rightarrow{n}=\frac{\mathrm{ln}\:\left(\frac{{A}}{{P}}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{r}\right)} \\ $$$${n}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{1}.\mathrm{1}}=\mathrm{7}.\mathrm{27}\:{years} \\ $$$$\left({b}\right) \\ $$$${with}\:{k}\:{compounding}\:{periods}\:{in}\:{one} \\ $$$${year}: \\ $$$${A}={P}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{{nk}} \\ $$$${continuous}\:{means}\:{k}\rightarrow\infty: \\ $$$${A}={P}×\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{{nk}} \\ $$$${A}={P}×\left[\underset{{k}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{r}}{{k}}\right)^{\frac{{k}}{{r}}} \right]^{{nr}} ={Pe}^{{nr}} \\ $$$$\Rightarrow{n}=\frac{\mathrm{ln}\:\left(\frac{{A}}{{P}}\right)}{{r}} \\ $$$${n}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{0}.\mathrm{1}}=\mathrm{6}.\mathrm{93}\:{years} \\ $$
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
$${thank}\:{you}\:{sir} \\ $$