Compound Interest Application

Step 1: Identify the known variables.

Remember that the rate must be in decimal form and n is the number of compoundings per year.

Since this situation has an annual interest rate there is only 1 compounding per year.

A = ? Account balance

P = $700 Starting value

r = 0.075 Decimal form

n = 1   No. compound.

t = 10 No. of years

Step 2: Substitute the known values.

$A=700{\left(1+\frac{0.075}{1}\right)}^{\left(1\right)\left(10\right)}$

Step 3: Solve for A.

$A=700{\left(1+\frac{0.075}{1}\right)}^{\left(1\right)\left(10\right)}$  Original

A = 700(1.075)¹⁰   Simplify

A = $1442.72  Multiply

Step 1: Identify the known variables.

Remember that the rate must be in decimal form and n is the number of compoundings per year.

Since this situation has quarterly compounding there are 4 compoundings per year.

A = $5046.02   Account balance

P = ?     Principal

r = 0.055   Decimal form

n = 4    No. compound.

t = 5    No. of years

Step 2: Substitute the known values.

$5046.02=P{\left(1+\frac{0.055}{4}\right)}^{\left(4\right)\left(5\right)}$

Step 3: Solve for P.

5046.02 = P(1.01375)²⁰  Original

$\frac{5046.02}{{1.01375}^{20}}=P$   Divide

P = $3840.00

Step 1: Identify the known variables.

Remember that the rate must be in decimal form and n is the number of compoundings per year.

Since this situation has bimonthly, twice a month, compounding there are 24 compoundings per year.

A = 4 x $2500     Account balance

P = $2500 Principal

r = 0.09 Decimal form

n = 24 No. compound.

t = ?   No. of years

Step 2: Substitute the known values.

$10,000=2500{\left(1+\frac{0.09}{24}\right)}^{\left(24\right)\left(t\right)}$

Step 3: Solve for t.

10,000 = 2500(1.00375)^24t   Original

4 = (1.00375)^24t   Divide

log_1.00375 4 = log_1.00375 (1.00375)^24t Log

log_1.00375 4 = 24t   Inverse

$\frac{lo{g}_{1.00375} 4}{24}=t$   Divide

$\frac{1}{24} \times \frac{log 4}{\log 1.00375}=t$ Change base

$t\approx 15.4$

Step 4: Solve for Ashton's age.

$5+15.4=20.4\approx 20$ years old