[tex]\$2000[/tex] will become [tex]\$4000[/tex] in 19.80 years that is approximately 20 years when compounded continuously at the annual interest rate of [tex]3.5\%[/tex]
Solution:
Given that
Amount investe by Annabell = [tex]\$2000[/tex],
Rate if interest [tex]= 3.5\% = 0.035[/tex]
[tex]Required \ amount = double \ of \ investment = 2\times \$2000 = \$4000[/tex]
And most important thing that interest is compounded continuously . Formula of Amount where interest is compounded continuously is as follows ,
[tex]\mathrm{A}=\mathrm{P} e^{\mathrm{r}{t}}[/tex]
Where A is final amount,
P is principal Amount,
r = rate of interest
And t = duration in years
In our case [tex]A = \$4000; \ P = \$2000; \ r = 0.035[/tex]
Need to evaluate t that is number of year.
On substituting given values in formula of amount we get
[tex]\begin{array}{l}{4000=2000 e^{0.035 t}} \\\\ {=>\frac{4000}{2000}=e^{0.035 t}} \\\\ {\Rightarrow 2=e^{0.035 t}}\end{array}[/tex]
Taking log both the sides,
[tex]\ln (2)=0.035 \mathrm{t} \times \ln (\mathrm{e})[/tex]
[tex]\Rightarrow \mathrm{t}=\frac{\ln (2)}{0.035}==19.80[/tex]
That is approximately 20 years.
