I am having a hard time solving this problem, Please Help!

Question

1. An account with $5000 compounded continuously has a rate of 10%.

a) Form and exponential function 𝐴 = 𝑃𝑒𝑟𝑡. A=_____________

b) Here is a table filled in using the function. Check a few of the years below to verify your function.

Time Amount Time (years) (dollars) (years)---Amount (dollars)

0 ---5000--- 8 --- 11127

1 ---5526 ---9 --- 12297

2--- 6107 ----10 ---13591

3 ----6749 ----11 ---15020

4--- 7459--- 12 ---16599

5--- 8243 ---13 ---18345

6 ---9110--- 14 ---20274

7--- 10068--- 15 ---22406

c) Find the time it takes the account to double and circle those times in the table above.

d) Use the formula 𝑡 = 𝑙𝑛(2) to find the exact time to double. Time to double:______________

𝑟

e) Using this doubling time, estimate what the account will be worth in 21 years? No calculator!

g) What will the account be worth in 30 years?

h) If you invest $5000 each year for 30 years, then it would be the same as adding the worth of each investment of $5000 in the account for 1 year, 2 years, ...., 29 years, and 30 years. Or, we could use the

formula: 𝐴 = 𝑃 (𝑒𝑟𝑇 − 1) to find the total amount in the account after 30 years of annual deposits of 𝑟

$5000. This would be a total deposit of $5000(30 years)=$150,000. But look at what it will become by using the formula to find A, the account balance.

With P=$5000, r=0.10 and T=30 plugged into the formula, Find A. A= 5000 (𝑒(0.1∙30 − 1) 0.1

A=$_____________________

Answer 1

A) P=Poe^rt for continuous compounding interest.

P= 5000e^0.1t

B) For t=0, P=5000e⁰=5000

for t=1, P=5000e^0.1=5526

for t=2, P=5000e^0.2=6107

Thus the formula is correct.

C) Time taken for the amount to double is approximately 7 years as seen from the table.

D) Exact time to double:

P=Poe^0.1t. P=2Po

2Po = Poe^.01t

t = ln(2)/0.1 = 6.93 years

E) Amount double in 6.93 years

after 6.93 years, amount 2*5000= 1000

after 2*6.93 years, amount 2*10000= 20000

after 3*6.92 years, amount 2*20000= 40000

ie= 20.79 years, amount = 40000

after 21 years, amount 40000+40000*(.21/6.93)

Estimated amount = 41212.12

Exact amount = 5000e^21*0.1 = 40830.85

G) amount in 30 years

= 5000e^20*0.1 = 100427.685

H) If 5000 is added each year:

A=P/r(e^rt-1)

5000/0.1(e^0.1*30-1)

=954276.845