Question

Hello I need help with parts b, c and d

Answer 1

(b)

`t\approx4.11\text{ years}`

(c) 14 years

(d) At t = 10 years, S = $18,123.77

At t = 20, S = $36,496.80

Step-by-step explanation:

b) We're asked to use the below graph to estimate when the future value will be $12,000;

Notice that each interval on the vertical axis represents 4,000 units and each interval on the horizontal axis represents 2 units.

Looking at the given graph, we can see that at S = $12,000 on the vertical axis, we can read off the intersection point on the horizontal axis from the graph and it is at t equals to approximately 4.11 years.

So the correction option is;

`t\approx4.11\text{years}`

c) When S = $23,980, let's go ahead and determine the value of t by substituting S with 23,980 and solving for t as seen below;

`\begin{gathered} S=9000e^(0.07t) \\ 23980.11=9000e^(0.07t) \\ (23980.11)/(9000)=e^(0.07t) \\ 2.6645=e^(0.07t) \\ \ln2.6645=lne^(0.07t) \\ \ln2.6645=0.07t \\ t=(2.6645)/(0.07) \\ t=14\text{ years} \end{gathered}`

Therefore, at S = $23.980.11, t = 14 years

d) When t = 10, let's go ahead and solve for S as seen below;

`\begin{gathered} S=9000e^(0.07t) \\ When\text{ t = 10,} \\ S=9000e^(0.07*10) \\ S=9000e^(0.7) \\ S=9000\left(2.0138\right) \\ S=18,123.77 \end{gathered}`

So at t = 10 years, S = $18,123.77

At t = 20, let's solve for S;

`\begin{gathered} S=9000e^(0.07t) \\ when\text{ t = 20} \\ S=9000e^(0.07*20) \\ S=9000e^(1.4) \\ S=9000\left(4.0552\right) \\ S=36,496.80 \end{gathered}`

Therefore, at t = 20, S = $36,496.80