Given: A woman invests $6300 in an account that pays 6% interest per year, compounded continuously.
Required: a) To determine the amount after 2 years.
b) To determine how long it will take for the amount to be $8000.
Explanation: The amount, A after t years with an interest rate of r is given by-
[tex]A=Pe^{rt}[/tex]
Here,
[tex]\begin{gathered} P=6300 \\ r=\frac{6}{100} \\ =0.06 \\ t=2 \end{gathered}[/tex]
Substituting the values, we get-
[tex]\begin{gathered} A=6300e^{0.06\times2} \\ =7103.23 \end{gathered}[/tex]
Hence the amount after 2 years is $7103.23
Next, let t be the time it takes for the amount to be $8000-
[tex]\begin{gathered} 8000=6300e^{0.06t} \\ \frac{8000}{6300}=e^{0.06t} \\ \ln(1.2698)=0.06t \end{gathered}[/tex]
Further solving for t as-
[tex]t=3.98\text{ years}[/tex]
Hence, it takes 3.98 years for the amount to be $8000.
Final Answer: a) $7103.23
b) 3.98 years
