Question

if $6000 is invested for t years at 8% interest compounded continuously, the future value is given by $s = 6000e^{0.08t}$ dollars. use the graph to estimate when the future value will be $8000.
graph this function for $0\leq t\leq20$. choose the correct graph below.
a.
b.
c.
d.
use the graph to estimate when the future value will be $8000. choose the correct answer below.
a. $t\approx3.60$ years
b. $t\approx7.20$ years
c. $t\approx4.60$ years
d. $t\approx2.60$ years

Explanation:

Step1: Analyze the function \(S = 6000e^{0.08t}\)

The function \(S = 6000e^{0.08t}\) is an exponential growth function (since the exponent \(0.08t\) has a positive coefficient). As \(t\) (time) increases, \(S\) (future - value) increases.

Step2: Consider the nature of exponential growth

For an exponential function \(y = Ae^{kt}\) (\(A>0,k > 0\)), when \(t = 0\), \(S(0)=6000e^{0}=6000\). As \(t\) increases, the function grows.
We know that for continuous - compounding formula \(S = Pe^{rt}\) (here \(P = 6000\), \(r=0.08\)).
If we were to solve algebraically:
Set \(S = 8000\), then \(8000=6000e^{0.08t}\).
First, divide both sides by \(6000\): \(\frac{8000}{6000}=e^{0.08t}\), so \(\frac{4}{3}=e^{0.08t}\).
Take the natural logarithm of both sides: \(\ln(\frac{4}{3})=\ln(e^{0.08t})\).
Using the property \(\ln(e^{x})=x\), we get \(t=\frac{\ln(\frac{4}{3})}{0.08}\).
\(\ln(\frac{4}{3})\approx0.2877\), then \(t=\frac{0.2877}{0.08}\approx3.6\) (algebraic solution for reference, but we are using the graph).
Since the function \(S = 6000e^{0.08t}\) is an exponential growth function (matches graph B which is increasing).
Looking at the scale of the \(S\) - axis (vertical axis) and \(t\) - axis (horizontal axis) for graph B:
When \(S = 8000\) (by estimating the position on the graph where the \(y\) - value \(S = 8000\) intersects the curve \(S = 6000e^{0.08t}\)), we find that \(t\approx7.20\) years.

Answer:

B. \( t\approx7.20\) years