5.1 Exponential Models and Laws
Key Takeaways
- An exponential model has a constant multiplier for equal input steps; growth uses a base greater than 1 and decay uses a base between 0 and 1.
- In y = a(b)^x, a is the value when x = 0 and b is the repeated multiplier, not the percent rate by itself.
- A vertical shift changes the horizontal asymptote of an exponential graph, while the domain usually remains all real numbers.
- The exponent laws are most useful when they help rewrite both sides with a common base before solving.
Why exponential models matter
Relations and Functions is the largest Math 30-2 reporting area, and exponential functions sit in the part of the course where algebra, graph reading, calculator work, and context all meet. The 2025-2026 Mathematics 30-2 information bulletin includes exponential functions on the formula sheet and notes that students often struggle when an exponential equation cannot be rewritten with a common base.
That is a useful warning: the diploma is not only asking whether you remember a formula. It is asking whether you can recognize the structure of repeated multiplication, decide what each parameter means, and communicate the result in the units of the problem.
An exponential model has the form y = a(b)^x, where a is the starting value when x = 0 and b is the common multiplier for each one-unit increase in x. If b > 1, the model shows growth. If 0 < b < 1, the model shows decay. The input x is in the exponent, so this is different from a polynomial such as y = 3x^2. On a graph, the basic model approaches a horizontal asymptote. Without a vertical shift, that asymptote is y = 0; with a model such as y = a(b)^x + c, it becomes y = c.
Reading the multiplier
A percent change must be converted to a multiplier before it belongs in the model. A 6% increase has multiplier 1.06, because the new amount is 100% + 6% of the old amount. A 6% decrease has multiplier 0.94, because the new amount is 100% - 6% of the old amount. A half-life model has multiplier 0.5 for each half-life. A doubling model has multiplier 2 for each doubling period.
| Situation | Multiplier | Model cue |
|---|---|---|
| Grows 12% each year | 1.12 | add the percent to 1 |
| Depreciates 18% each year | 0.82 | subtract the percent from 1 |
| Doubles every 5 days | 2 per 5-day interval | use x/5 if x is days |
| Halves every 3 hours | 0.5 per 3-hour interval | use x/3 if x is hours |
Exponent laws you actually use
The formula sheet and calculator can help, but you still need clean exponent rules. The most common laws are b^m * b^n = b^(m+n), b^m / b^n = b^(m-n), (b^m)^n = b^(mn), b^0 = 1 when b is not 0, and b^(-n) = 1 / b^n. In diploma work, these laws usually support one of two moves: simplify a model or rewrite an equation with matching bases. For example, 3(2)^(x+1) = 48 becomes 2^(x+1) = 16, then 2^(x+1) = 2^4, so x + 1 = 4 and x = 3.
Worked example: depreciation
A snow-removal machine is purchased for $12,000 and loses 18% of its value each year. The model is V(t) = 12000(0.82)^t, where t is the number of years after purchase. After 4 years, V(4) = 12000(0.82)^4, which is about $5,424. The base 0.82 says the machine keeps 82% of its value each year, not that it loses 82%.
Now read the same model graphically. The y-intercept is 12000. The graph decreases quickly at first and then flattens as it approaches y = 0. In the real context, the machine will not literally have negative value, so the model is only reasonable over a practical time window. If a written-response part asks for interpretation, say that the value is predicted by a depreciation model and give the units, not just the number.
Diploma traps
The most reliable way to identify an exponential table is to compare ratios between consecutive y-values for equal x-steps. Constant first differences suggest a linear model. Constant second differences suggest a quadratic model. Constant ratios suggest an exponential model. On numerical-response items, keep full calculator values until the final requested rounding. On multiple-choice items, check whether the answer uses the percent rate, such as 0.18, instead of the decay multiplier, such as 0.82.
A second trap is comparing bases without reading the coefficient. For y = 500(1.04)^x and y = 300(1.08)^x, the second model grows faster, but the first model starts higher. A question about the value at x = 0, the long-term growth rate, or the time when one overtakes the other may have different answers. Always identify what the question is asking before calculating.
Calculator and step-size checks
A graphing calculator is useful, but it will not decide the model for you. When x-values increase by more than 1, adjust the multiplier to match the step size. If a value grows from 900 to 1152 over 3 years, the 3-year multiplier is 1152/900 = 1.28. The annual multiplier is 1.28^(1/3), about 1.086, so the annual growth rate is about 8.6%, not 28%. Diploma distractors often use the total change across the whole interval as if it were the one-step rate.
For numerical-response work, estimate first. If the base is greater than 1, later outputs should be larger unless a negative coefficient or vertical shift changes the context. If the base is between 0 and 1, later outputs should move toward the horizontal asymptote. This quick direction check catches many entry and window mistakes before you commit an answer.
A population is modeled by P(t) = 6400(1.035)^t, where t is measured in years. What does 1.035 represent?
Which table is most consistent with an exponential model for equal x-steps?
A substance has amount A(h) = 80(0.5)^(h/6), where h is measured in hours. What is the best interpretation?