Question

Odette has two investments that she purchased at the same time. Investment 1 cost $10,000 and earns 4% interest each year. Investment 2 cost $8000 and earns 6% interest each year.

Write exponential growth functions

that could be used to find v1(t) and v2(t), _________________________________

the values of the investments after t years.


Find the value of each investment after 5 years.

Explain why the difference between their values, ____________________________ which was initially $2000, is now less

Answer 1

Answer: The functions are v1(t)=10000*(1.04)^5 and v2(t)=8000(1.06)^5

v1(5)=12166$

v2(5)=10705 $

Explanation:

Given:

Investment cost for 1st=10,000$ ,at 4%rate.

Investment cost for 2nd=8000$ ,at 6% rate.

To Find:

Exponential growth functions ,

that could be used to find v1(t) and v2(t),

the value of each investment after 5 years

And difference between their values.

Solution:

Consider for 1st investment function be v1(t) and for 2nd be v2(t)

and t be time period

Exponential growth is given by, as time function,

v(t)=initial cost*(1+rate)^time period.

for 1st investment its cost is 10000$ at 4 % rate.

`v1(t)=10000(1+0.04)^t`

`v1(t)=10000(1.04)^t`

for 2nd investment its cost is 8000$ at 6% rate.

`v2(t)=8000(1+0.06)^t`

`v2(t)=8000(1.06)^t`

For t=5 calculating for both investment,

`v1(5)=10000(1.04)^5`

`=10000*1.2166`

v1(5)=12166 $

For 2nd investment,

`v2(5)=8000(1.06)^5`

`=8000*1.3382`

v2(t)=10705.8 $

Now after 5 years the difference is =v1(t)-v2(t)

Which is about 1460 $ only.

This because of the both functions depends upon time,initial investment and rate of interest .

Here time period being same.But the rate of interest are different and initial values are different.