Question

Suppose you deposit $2000 in a savings account that pays interest at an annual rate of 4%. If no money is added or withdrawn from the account, answer the following questions.
A. how much will be in the account after three years?
B.How much will be in the account after 18 years?
C. How many years will it take for the account to contain $2500?
D. How many years will it take for the account to contain $3000?

Answer 1

The formula for compound growth is:

$P=P_{0}(1+\frac{r}{n})^{nt}$

Where,

• P is the future value
• $P_0$ is the initial desposit
• r is the annual rate of interest
• n is the number of times compounding happens ( n = 1 when annual compounding, n = 2 when semi-annual compounding, n = 4 when quarterly compounding etc.)
• t is time in years

A.

Here we want to know P, given $P_{0}=2000$, r is 0.04, n is 1 (since annual compounding) and t is 3.

$P=2000(1+0.04)^{3}\\=2000(1.04)^{3}\\=2249.73$ dollars

B.

Everything is same in this problem as part A, just t is not 3, t is 18. Similarly, we solve:

$P=2000(1+0.04)^{18}\\=2000(1.04)^{18}\\=4051.63$ dollars

C.

Here we want to figure out t, given P=2500, $P_{0}=2000$, r is 0.04, n is 1 (since annual compounding). We have:

$2500=2000(1+0.04)^{t}\\2500=2000(1.04)^{t}\\\frac{2500}{2000}=1.04^{t}\\1.25=1.04^{t}$

To solve this exponential part, we take Natural Logarithm (ln) and use our properties of logarithms and solve for t. [the property we are going to use is $ln(a)^{n}=n*ln(a)$

$ln(1.25)=ln[1.04^{t}]\\ln(1.25)=t*ln(1.04)\\t=\frac{ln(1.25)}{ln(1.04)}\\t=5.69$ years

D.

It is similar to part C, only P=3000 instead of P=2500. Let us setup the equation and solve for t.

$3000=2000(1+0.04)^{t}\\\frac{3000}{2000}=(1.04)^{t}\\\frac{3}{2}=(1.04)^{t}\\ln(\frac{3}{2})=ln[{1.04}^{t}]\\ln(\frac{3}{2})=t*ln(1.04)\\t=\frac{ln(\frac{3}{2})}{ln(1.04)}\\t=10.34$ years

ANSWER:

A. $2249.73

B. $4051.63

C. 5.69 years

D. 10.34 years

Answer 2

Hi there
The formula is
A=p (1+r)^t
A future value
P present value
R interest rate
T time

A) A=2,000×(1+0.04)^(3)=2,249.728

B) A=2,000×(1+0.04)^(18)=4,051.63

C) 2500=2000 (1+0.04)^t
Solve for t
T=log(2,500÷2,000)÷log(1+0.04)
T=5.7 years

D) t=log(3,000÷2,000)÷log(1+0.04)
t=10.3 years

Hope it helps