Exponential Equations: Continuous Compound Interest Application
The formula for continuously compounded interest, which is different from the compounded interest formula, is:
COMPOUND INTEREST FORMULA
A = Pe^{rt}
Where A is the account balance, P the principal or starting value, e the natural base or 2.718, r the annual interest rate as a decimal and t the time in years.
Let's solve a few continuously compounded interest problems.
Step 1: Identify the known variables. Remember that the rate must be in decimal form. |
A = ? Account balance P = $2750 Starting value r = 0.0725 Decimal form t = 15 No. of years |
Step 2: Substitute the known values. |
A = Pe^{rt} A = 2750e^{(0.0725)(15)} |
Step 3: Solve for A. |
A = 2750e^{1.0875} Original A = $8129.36 Simplify |
Step 1: Identify the known variables. Remember that the rate must be in decimal form. |
A = $12,750 Account balance P = $5000 Principal r = ? Decimal form t = 10 No. of years |
Step 2: Substitute the known values. |
A = Pe^{rt} 12,750 = 5000e^{(r)(10)} |
Step 3: Solve for r. |
12,750 = 5000e^{10r} Original 2.55 = e^{10r} Divide by 5000 ln 2.55 = ln e^{10r} Take ln ln 2.55 = 10r Inverse $\frac{\mathrm{ln}2.55}{10}=r$ Divide by 10 $0.094\approx r\approx 9.4\%$ |
Step 1: Identify the known variables. Remember that the rate must be in decimal. |
A = $13,700 Account balance P = ? Principal r = 0.083 Decimal form t = 4 No. of years |
Step 2: Substitute the known values. |
A = Pe^{rt} 13,700 = Pe^{(0.083)(4)} |
Step 3: Solve for P. |
13,700 = Pe^{0.33} Original $\frac{13,700}{{e}^{0.33}}=P$ Divide by e^{0.33} $9816.48 = P |
Related Links: Math algebra Exponential Equations: Exponential Growth and Decay Application Exponential Equations: Introduction and Simple Equations Algebra Topics Exponential Functions |
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