The same exponential model can be written in different ways, and each way answers a different question. The tool is the power-of-a-power rule, applied to exponential models.
Why it works
Since $(b^m)^t = b^{mt}$, you can regroup an exponent. Take $A = 100 \cdot 1.21^t$ with $t$ in years. Because $1.21 = 1.1^2$, we get $A = 100 \cdot (1.1^2)^t = 100 \cdot 1.1^{2t}$. Now $2t$ counts half-years, so the amount grows by $10\%$ every half year. Going the other way, a monthly factor $m$ gives a yearly factor $m^{12}$, because a year holds $12$ months of growth.
| Form | What it shows |
|---|---|
| $100 \cdot 1.21^t$ | $21\%$ growth per year |
| $100 \cdot 1.1^{2t}$ | $10\%$ growth per half year |
| $100 \cdot \left(1.21^{1/12}\right)^{12t}$ | about $1.6\%$ growth per month, since $1.21^{1/12} \approx 1.016$ |
Key idea
Use $b^{mt} = (b^m)^t$ to change the time unit. A rate for a shorter period is found with a root, $b^{1/n}$, never by dividing the percent by $n$.
Worked examples
Worked example: Half-year factor
Rewrite $A = 50 \cdot 1.44^t$ to show the half-year growth rate.
- $1.44 = 1.2^2$, so $A = 50 \cdot (1.2^2)^t = 50 \cdot 1.2^{2t}$.
- The factor per half year is $1.2$.
Answer: $20\%$ growth every half year.
Worked example: Monthly to yearly
A balance grows $1\%$ per month. What is the yearly growth rate?
- In $t$ years there are $12t$ months: $B = B_0 \cdot 1.01^{12t} = B_0 \cdot (1.01^{12})^t$.
- $1.01^{12} \approx 1.1268$.
Answer: About $12.68\%$ per year, a little more than $12\%$.
Common mistake
Dividing a yearly rate by $12$ to get a monthly rate. $12\%$ a year is not $1\%$ a month: $1\%$ a month gives $1.01^{12} \approx 1.127$, which is $12.7\%$ a year, because each month earns on the earlier months' growth. Check a monthly factor by raising it to the $12$th power; you should get the yearly factor back.