Exponential Growth and Decay

Linear change adds the same amount each step. Exponential change multiplies by the same amount each step. That one-word difference, adds versus multiplies, is what this entire topic rests on, and it is what the test is really checking when it asks whether a savings account, a population, a bacteria culture, or a depreciating car is best modeled by a linear or an exponential function.

The SAT places exponential functions in the Advanced Math domain and loves context: compound interest, populations, medicine leaving the bloodstream, radioactive decay, and the resale value of equipment. Many of these questions never ask you to compute a final number at all. Instead they ask you to pick the correct model from four expressions, which means the tested skill is converting a stated percent change into a multiplier and putting it in the right place. The ACT asks similar questions and adds more direct evaluation, such as finding a term of a geometric sequence or computing a value after a given number of half-lives.

Students who struggle here almost always struggle in the same specific way: they put the percent itself into the base. A quantity that grows 6 percent per year is not multiplied by 0.06 each year, because that would shrink it to almost nothing. It is multiplied by 1.06, because it keeps everything it had and adds 6 percent more. Getting that translation automatic is worth more points than any other single habit in this lesson.

The exponential model is y = a times b^x, where a is the starting amount, the value when x = 0, because b⁰ = 1 for any nonzero b. The number b is called the base, the growth factor, or the multiplier, and it is the number the quantity is multiplied by for each one-unit increase in x. That interpretation follows directly from the algebra: y at x + 1 equals a times b^(x + 1), which equals b times a times b^x, which is b times y at x. Every step forward multiplies by b, no matter where you start.

Compare that to a linear function y = mx + c, where every step forward adds m. This produces the diagnostic test used on every table question. For a linear relationship, the first differences between consecutive outputs are constant. For an exponential relationship, the ratios between consecutive outputs are constant. Given a table with equally spaced inputs, subtract to test for linear and divide to test for exponential. If neither is constant, the relationship is neither.

When b > 1 the function grows, and when 0 < b < 1 the function decays. The reason decaying functions never reach zero is that multiplying a positive number by a positive fraction always leaves something positive, however small; graphically the curve approaches the horizontal axis as an asymptote but never touches it. This is a genuine conceptual difference from a linear decline, which crosses zero and keeps going negative, and the test uses it: a quantity that loses a fixed percent of what remains each year behaves very differently from one that loses a fixed amount each year.

To turn a percent change into a multiplier, write b = 1 + r, where r is the percent change written as a decimal, positive for an increase and negative for a decrease. An 8 percent yearly increase gives b = 1 + 0.08 = 1.08. An 8 percent yearly decrease gives b = 1 - 0.08 = 0.92. A 150 percent increase gives b = 1 + 1.50 = 2.50. Why the 1 is there is worth stating plainly: the quantity keeps 100 percent of itself and then gains or loses r on top, so the multiplier is 100 percent plus r, which is 1 + r in decimal form. A useful check is that a decay multiplier must land strictly between 0 and 1, so if you compute a decay base above 1 you have made a sign error.

The exponent must be counted in the same time unit as the rate. If a rate is quoted per year but time is measured in months, the model becomes y = a times b^(t/12), because 12 months make one compounding period. This is exactly the structure of doubling time and half-life problems. A quantity that doubles every d units satisfies y = a times 2^(t/d), and a quantity that halves every h units satisfies y = a times (1/2)^(t/h). To see why, notice that at t = d the exponent is 1, so the value is 2a; at t = 2d the exponent is 2 and the value is 4a. The shortcut for clean numbers is to count how many whole doubling or halving periods fit into the elapsed time and multiply or divide accordingly, so three half-lives means dividing by 2 three times, that is dividing by 8, not by 6. Compound interest is the same idea with named parts: A = P(1 + r/n)^(nt), where P is the principal, r is the annual rate as a decimal, n is the number of compounding periods per year, and t is the number of years. The base 1 + r/n is the multiplier for one period and nt counts the periods, so with annual compounding n = 1 and the formula collapses to P(1 + r)^t.

Finally, learn to recognize the question types. Phrases like increases by a constant percent, is multiplied by, decreases by half every, compounded annually, and each year the population is p percent larger all signal exponential models. Phrases like increases by a constant amount, per additional unit, and at a constant rate of change signal linear models. When a question gives a table and asks which type of function best models the data, do both tests, differences and ratios, before you commit; when it asks you to select an expression rather than a value, check the base first, then the exponent, then the coefficient, because that is the order in which the wrong answers are usually built.

In the interactive graph below, start from y = 2 times 1.5^x and drag b. Watch the curve flatten as b approaches 1, flip into decay the moment b drops below 1, and notice that the y-intercept marker never moves while you do it, because a times b⁰ is always a. Then drag a and see the intercept rise and fall while the shape stays the same, and finally drag k to lift the dashed asymptote off the x-axis; the whole curve follows it, which is exactly why a quantity that decays toward some floor levels off at k rather than at zero.

Putting the raw percent in the base. A 6 percent annual increase means multiplying by 1.06, not by 0.06 and not by 6. Writing 4000 times 0.06^t describes a quantity that loses 94 percent of itself every year. Say the sentence in words before you write the formula: it keeps all of what it had, plus six percent more, so the multiplier is one point zero six.

Applying a percent change linearly. Losing 15 percent per year for three years is not losing 45 percent, because the second year's 15 percent is taken from a smaller amount than the first year's. The correct multiplier is 0.85³ = 0.614125, a total loss of about 38.6 percent rather than 45 percent. Any time a problem says of the remaining amount or per year on the current value, the change compounds.

Miscounting doubling or halving periods. After three half-lives a quantity is divided by 2 three times, that is divided by 8, not divided by 6. Similarly, four doublings multiply by 16, not by 8. Write out the chain of values, such as 200, 100, 50, 25, when the numbers are small; it takes five seconds and eliminates the error.

Ignoring a mismatch between the rate's time unit and the exponent's time unit. If a population grows 30 percent per decade and the question asks about t years, the model is P = a times 1.3^(t/10), not 1.3^t. Ask yourself what value of t should make the exponent equal 1, and build the exponent so that it does.

Declaring a table exponential after checking only two rows. Two rows always have a ratio, so one ratio proves nothing. Check every consecutive pair, and confirm first that the inputs are equally spaced, because a constant ratio across unequal input steps does not give a single base.