Section 6.7: solutions
#1-4: Use the compound interest formula 𝐴 = 𝑃 (1 + 𝑟 𝑛
) 𝑛𝑡
to answer the following.
1) An initial deposit of $1,000 earns 4% interest compounded twice per year. How much will be in the account after 5 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the answer.
A
Solve
P
1000 r
.04 n
2 t
5
𝐴 = 1000 (1 +
.04
2
)
2∗5
(note: you must put the 2*5 in a parenthesis when you use your calculator)
Answer: The investment will be worth $1,218.99 in 5 years
3) An initial deposit of $15,000 earns 2% interest compounded quarterly. How much will be in the account after 8 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the answer.
A
Solve
P
15000 r
.02 n
4 t
8
𝐴 = 15000 (1 +
.02
)
4
4∗8
(note: you must put the 4*9 in a parenthesis when you use your calculator)
Answer: The investment will be worth $17,595.65 in 8 years.
#5-8: Use the formula A=Pe rt to answer the following.
5) An initial investment of $5,000 earns 6% interest compounded continuously. What will the investment be worth in 5 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the answer.
A
Solve
A = 5000e .06*5
P
5000 r
.06 t
5
Answer: The investment will be worth $6,749.29 in 5 years.

7) An initial investment of $15,000 earns 3% interest compounded continuously. What will the investment be worth in 6 years?
I need to solve for A. I will put the following values in the formula, then use my calculator to get the answer.
A
Solve
A = 15000e .03*6
P
15000 r
.03 t
6
Answer: The investment will be worth $17,958.26 in 6 years.
9) How long will it take an initial investment of $1,000 to triple if it is expected to earn 6% interest compounded continuously? (Round to 1 decimal place)
I will use this formula because interest is compounded continuously: A=Pe rt
The $1,000 is the P, the A is triple this amount so the A is $3,000. I need to solve for t.
I will put these values in the formula and solve for t.
A
3000
P
1000 r
.06
3000 = 1000e .06t
(divide both sides by 1000)
3 = e .06t
(take ln of each side)
Ln 3 = ln e .06t
(use power to product rule)
Ln 3 = .06tln e (lne = 1, so drop it)
Ln 3 = .06t (divide by .06 then use calculator) ln 3
= 𝑡
.06
Answer: 18.3 years (rounded to one decimal place as required) t solve

11) How long will it take an initial investment of $100,000 to grow to $1,000,000 if it is expected to earn
4% interest compounded continuously? (Round to 1 decimal place)
I will use this formula because interest is compounded continuously: A=Pe rt
The $100,000 is the P, the A is$1,000,000. I need to solve for t.
I will put these values in the formula and solve for t.
A
1,000,000
P
100,000 r
.04
1,000,000 = 100,000e .04t
(divide both sides by 10,000)
10 = e .04t
(take ln of each side)
Ln 10 = ln e .04t
(use power to product rule)
Ln 10 = .04tln e (lne = 1, so drop it)
Ln 10 = .04t (divide by .04 then use calculator) ln 10
= 𝑡
.04
Answer: 57.6 years (rounded to one decimal place as required) t solve
13) What will a $200,000 home cost in in 5 years if the price appreciation over that period is expected to be 3% compounded annually?
I will use the 𝐴 = 𝑃 (1 + 𝑟 𝑛
) 𝑛𝑡
formula because the growth is compounded annually. This tells me that n = 1
I am solving for A.
A
Solve
P
200,000
𝐴 = 200,000 (1 +
.03
)
1
1∗5 r
.03
Answer: The house will be cost $231,854.81 in 5 years. n
1 t
5

15) I will use the formula 𝐴 = 𝑃 (1 + 𝑟 𝑛
) 𝑛𝑡
as the growth in compounded annually. In this case n will equal 1.
I am solving for A:
A
Solve
P
100000
A = 100000 (1 +
.02
)
1
1∗3
= 106120.80
r
.02 n
1 t
3
Answer: The house will cost $106,120.80 in three years
17) I will use the formula 𝐴 = 𝑃 (1 + 𝑟 𝑛
) 𝑛𝑡
as the growth in compounded annually. In this case n will equal 1.
I am solving for A:
A
Solve
P
80
A = 80 (1 +
.04
)
1
1∗5
= 97.33
r
.04
Answer: The tuition will be $97.33 in 5 years n
1 t
5