1. Given the modeling function, $A(t) = 100e^{0.02t}$ that depends on time, $t$ years, for $\$100$ invested at 2% per annum compounded continuously, find  the value of the investment  after 2 years, that is find $A(2)$. [Evaluate the function]
   
2. Given the investment modeling function, $A(t) = 100e^{0.02t}$ find the time, $t_v$, where the value of the investment will be $\$200$, i.e., when  $A(t_v) = 200$. [Solve the equation]
3. Suppose  $A(t)=Pe^{rt}$ and the value of the investment was $\$120$ after one year and $\$130$ after two years. Find the principal and the rate of interest for the investment.

1. Given  $A(t) = 100e^{0.02t}$, find $A(2)$. Evaluate the function: $A(2) = 100e^{0.02\cdot 2} = 100e^{0.04}  \approx 104.081$, so the value of the investment after $2$ years is about $\$104.08$.

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2. Given $A(t) = 100e^{0.02t}$,  solve the equation  $A(t_v) = 100e^{0.02t_v} = 200$. This gives $e^{0.02t_v} = 2$;  $0.02 t_v =\ln(2)$ and $t_v = \frac {\ln(2)}{0.02} \approx 34.6573$. It will be approximately 34 years and 8 months before the value of the investment will be \$200.

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3. We suppose  $A(t)=Pe^{rt}$.  and $A(1) = Pe^r =120$  while $A(2) = Pe^{2r} = 130$ . Thus $\frac{A(2)}{A(1)} = \frac {130}{120} = e^r$.
   Thus $r = \ln(\frac {13}{12}) \approx 0.080$ or about $8\%$.
   The principal, $P$, is the initial investment, and $P = \frac {120}{e^r} = \frac {120}{\frac {13}{12}} = \frac {1440}{13} \approx 110.769$.
   So the principal was about $\$110.77$.

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