- 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]

- 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] - 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.

- 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 $\$104.08$.

- 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.
- 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 $\$110.77$.

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 $C$.