SAT Exponential Growth: Initial Values, Factors, and Time Intervals

An exponential model repeatedly multiplies a starting amount by the same factor over equal time intervals. To read one, identify the amount at time zero, the multiplier, and the length of the interval that multiplier describes. A percent growth rate is only the added fraction; the multiplier also includes the amount already present.

College Board includes exponential equations and nonlinear functions in SAT Advanced Math. Its math skill descriptions include using exponential functions to model relationships. All examples below are original Kibo practice, not official exam items.

Read the starting value, factor, and interval

For a positive starting quantity, use $Q(t)=a b^{t/k}$, where $a>0$, $b>0$, and $k>0$.

• $a$ is the amount at $t=0$: $Q(0)=a b^0=a$.
• $b$ is the multiplier applied over each interval of length $k$.
• $t/k$ counts those intervals. The units of $t$ and $k$ must agree.

Growth uses $b=1+r$, with $r$ written as a decimal. Thus $20\%$ growth gives $b=1.20$. For a decrease of $r$, use $b=1-r$; a $15\%$ decrease gives $b=0.85$.

With a positive starting amount, $b>1$ means growth, $0<b<1$ means decay, and $b=1$ gives a constant amount. These readings apply to the displayed model; an added constant or shifted exponent requires evaluating the whole expression. OpenStax’s exponential-functions chapter develops the initial-value and constant-factor interpretation.

Worked example: why the increase gets larger

A design studio models its stored files as starting at $125$ gigabytes and increasing by $20\%$ each month. Let $m$ count completed months from that starting point. Then $S(m)=125(1.20)^m$.

The first increase is $0.20(125)=25$. The next increase is $0.20(150)=30$, because the percentage now applies to the larger amount. Check the ratios: $150/125=180/150=216/180=1.20$.

Adding $25$ each month would instead give the linear model $L(m)=125+25m$. At month two, $L(2)=175$, while $S(2)=180$. The two models agree at the first checkpoint but describe different rules.

The storage rule is a hypothetical planning assumption. Real data would need checking before using it as a forecast. If converting the percentage is the difficult step, review the percent-change guide before returning to repeated growth.

Worked example: the exponent counts growth periods

A lab model describes a culture’s biomass by $M(h)=48(1.5)^{h/2}$, measured in milligrams, where $h$ is hours since the starting measurement.

The starting mass is $48$ milligrams. The factor $1.5$ means $50\%$ growth over two hours, because $h=2$ makes the exponent $1$. After six hours, there have been $6/2=3$ growth periods:

$$M(6)=48(1.5)^3=162$$

Check by repeated multiplication: $48$, $72$, $108$, then $162$ milligrams at hours $0$, $2$, $4$, and $6$. Using an exponent of $6$ would apply six two-hour increases, reaching the model’s twelve-hour value.

Worked example: find a rate from spaced observations

Suppose a positive quantity follows $A(t)=a b^t$, with $t$ measured in months. You are given $A(2)=242$ and $A(4)=292.82$. What are the monthly growth rate and the starting value?

The ratio across the two-month gap is $292.82/242=1.21$. Therefore $b^2=1.21$, so the positive monthly factor is $b=1.10$. The monthly growth rate is $10\%$.

Work backward two months to find $a=242/(1.10)^2=200$. The model is $A(t)=200(1.10)^t$. Substitution checks both observations: $A(2)=242$ and $A(4)=292.82$.

The $21\%$ increase describes the entire two-month interval. Halving it to get $10.5\%$ per month would be incorrect: $(1.105)^2=1.221025$, which gives a different two-month factor.

The problem states that the relationship is exponential. Two measurements alone would not establish that a real process follows this rule at every other time.

Three original practice questions

1. For $P(d)=320(1.075)^d$, where $d$ is days elapsed, identify the initial value, daily multiplier, and daily percent growth rate.
2. A sample starts with $240$ milligrams of material. A model says the amount decreases by $15\%$ per day. Write the model and calculate the amount after two days.
3. An exponential model $G(w)=a b^w$, with positive $a$ and $b$, satisfies $G(1)=90$ and $G(3)=129.6$. Find the weekly multiplier $b$ and initial value $a$, where $w$ is weeks elapsed.

A short check before choosing an answer

• Locate time zero. Evaluate the model there; do not assume the first number in a table is the initial value.
• Separate the rate and multiplier. Include the amount retained when forming the multiplier.
• Label the interval. A ratio between observations describes their entire time gap. Keep the units beside your calculation.
• Check the rule with a second step. Repeated multiplication should reproduce the next stated value. Keep exact values until a requested final rounding step.

When an answer is wrong, record whether the error came from the starting point, the multiplier, or the exponent. The Kibo practice-error log provides a template for explaining the correction and checking a new problem later.

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Prepared with AI assistance. Original examples and practice answers were checked with separate exact calculations; no human review is claimed. Sources checked November 10, 2026. SAT is a registered trademark of the College Board, which is not affiliated with or endorsing this guide.