Example  ELF.APP.1.  Continuously Compound Interest.  $A(t)=Pe^{rt}$. $P=$ principal; $r =$ rate of interest per annum as decimal; $t =$ time invested.
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.

Solutions:
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 $\$104.08$. 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.
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 $\$110.77$. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com Comments: You can move the (red) point labelled x on the left axis of the mapping diagram to a position where the arrow head points to$f(x) = 20$and the corresponding point on the graph of$f$will move to the position where the graph of$f$crosses the line$ y = 20$. Check the box and the diagram will show the solution on both the mapping diagram and the graph. You can use the sliders to investigate other examples by changing the base,$A$, and$k$, as well as$C\$.