Exponential Equations: Compound Interest Application
The formula for compound interest is:
COMPOUND INTEREST FORMULA
$A=P{\left(1+\frac{r}{n}\right)}^{nt}$
Where A is the account balance, P the principal or starting value, r the annual interest rate as a decimal, n the number of compoundings per year and t the time in years.
Let's solve a few compound interest problems.
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 $\frac{1}{24}\times \frac{log4}{\mathrm{log}1.00375}=t$ Change base $t\approx 15.4$ |
Step 4: Solve for Ashton's age. |
$5+15.4=20.4\approx 20$ years old |
Related Links: Math algebra Exponential Equations: Continuous Compound Interest Application Exponential Equations: Exponential Growth and Decay Application Algebra Topics Exponential Functions |
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