Exponential models show up on the SAT as population growth, compound interest, depreciation, and decay. The questions ask for the percent rate, the starting amount, or the value after some time. All of them come down to reading the base of the exponent correctly and knowing that it is a multiplier, not an addition.
In a times b to the power t, a is the starting amount, and subtracting 1 from b gives the percent change per period.
The standard form is f of t equals a times b to the power t, where a is the value at time zero and b is the growth factor per period. If b is greater than 1 the quantity is growing, and the percent increase is b minus 1 written as a percent. If b is between 0 and 1 the quantity is shrinking, and the percent decrease is 1 minus b.
So in f of t equals 500 times 1.08 to the power t, the starting amount is 500 and the growth rate is 8 percent per period, since 1.08 minus 1 is 0.08. In g of t equals 900 times 0.85 to the power t, the starting amount is 900 and the quantity decreases by 15 percent per period, since 1 minus 0.85 is 0.15. Building the model from words runs the same way in reverse: a 3 percent annual decline gives a base of 0.97.
Exponential and linear behave very differently and the SAT tests the contrast. Linear adds the same amount each period. Exponential multiplies by the same factor each period, so the amount added grows as the total grows. In a table, constant differences mean linear and constant ratios mean exponential. Checking two consecutive ratios settles it.
The exponent is not always plain t. A model might read 2000 times 1.05 to the power t over 4, which means the quantity grows 5 percent every 4 years rather than every year. Or 300 times 2 to the power t over 10, which is a doubling every 10 of whatever units t measures. Divide the elapsed time by the period length to get the exponent, and read the units in the question rather than assuming years.
One structural check keeps the whole family straight. Substitute t equals zero. Any base to the zero power is 1, so the function collapses to the coefficient in front, confirming that the coefficient is the starting value. Then substitute t equals 1 to see one period of change and confirm that your percent matches the direction the situation describes.
Treating the growth as additive and adding the same amount each period, instead of multiplying by the growth factor each period.
A population doubles every 3 years. If it is 1000 now, what will it be in 12 years?
A patient receives 500 milligrams of a drug, and each hour 8% of the drug still present is cleared. To the nearest milligram, how much remains after 6 hours?
Why: Each period multiplies by 1 - 8/100 = 0.92, so the amount is 500(0.92)^6 ≈ 303. The value 460 applies the decrease a single time, which leaves far too much. Answering 330 means you counted one period too few.
For growth, add the percent as a decimal to 1, so a 6 percent increase gives a base of 1.06. For decay, subtract it from 1, so a 6 percent decrease gives a base of 0.94. Multiply that base raised to the number of periods by the starting amount.
That number is the length of one growth period. An exponent of t over 5 means the model multiplies by the base once every 5 time units. Set the exponent to 1 to find how long one full period takes.