A repeated percentage change is a geometric model. Use the rate per compounding period and count the same periods in the exponent. **Situation** | **Model** | **Meaning** Compound interest | A=P(1+ p100m)mt | P: initial deposit; p: nominal annual percentage rate; m: periods per year; t: years. Annual depreciation | V=P(1- d100)t | d: percentage lost each year; 0<d<100. Interest earned | I=A-P | Subtract the initial deposit from the accumulated amount. Example At 6% nominal annual interest compounded quarterly, the period rate is 1.5%. A deposit of $2000 grows over two years to A=2000(1.015)8 $2252.99. A machine worth $12000 depreciates by 15% each year. After three years, V=12000(0.85)3=$7369.50. Technology check In a financial solver, match payments per year and compounding periods to the stated model. For a single deposit with no payments, set the payment to zero. Use opposite signs for money paid out and money received; check the solver result by substituting into the formula. To find a time or rate, rearrange the same model. For A=Pqn, n= ln(AP)ln q, where q>0 and q 1. For annual crediting, test the next integer year when a target must first be reached.

A repeated percentage change is a geometric model. Use the rate per compounding period and count the same periods in the exponent.

$A=P\left(1+\dfrac{p}{100m}\right)^{mt}$

$P$: initial deposit; $p$: nominal annual percentage rate; $m$: periods per year; $t$: years.

$d$: percentage lost each year; $0<d<100$.

Subtract the initial deposit from the accumulated amount.

At $6\%$ nominal annual interest compounded quarterly, the period rate is $1.5\%$. A deposit of $\$2000$ grows over two years to $A=2000(1.015)^8\approx\$2252.99$.

A machine worth $\$12000$ depreciates by $15\%$ each year. After three years, $V=12000(0.85)^3=\$7369.50$.

In a financial solver, match payments per year and compounding periods to the stated model. For a single deposit with no payments, set the payment to zero. Use opposite signs for money paid out and money received; check the solver result by substituting into the formula.

To find a time or rate, rearrange the same model. For $A=Pq^n$, $n=\dfrac{\ln\left(\dfrac AP\right)}{\ln q}$, where $q>0$ and $q\ne1$. For annual crediting, test the next integer year when a target must first be reached.

Standard Form and Compound Interest

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Compound interest and depreciation

Compound interest and depreciation