Exponential Growth and Decay Models

8.2: Exponential Growth and Decay Models. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 2 of 63 parts Practice Flashcards

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Growth factors

Objective: Evaluate a repeated growth model.

Prerequisite skills: percent multipliers and powers. A rate r expressed as a decimal gives multiplier 1+r for each period. Multiply the starting amount by that factor once per elapsed period. The model assumes the same proportional rate repeatedly. Common error: multiplying by r alone and losing the original amount. Restrictions and interpretation: A growth factor is 1+r and a decay factor is 1−r; use a consistent time unit.

Formulas

$A(t)=A₀(1+r)^t$

Worked examples

Try itA population begins at 250 and grows 50% per year. What does the model predict after 2 years?Show answer

The growth factor is 1+0.50=1.5. After two years the model is 250(1.5)²=250×2.25=562.5. Therefore the requested answer is 562.5.

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Growth factors

Objective: Evaluate a repeated growth model.

Formulas

$A(t)=A₀(1+r)^t$

Worked examples

Try itA population begins at 250 and grows 50% per year. What does the model predict after 2 years?Show answer

The growth factor is 1+0.50=1.5. After two years the model is 250(1.5)²=250×2.25=562.5. Therefore the requested answer is 562.5.

Decay factors

Objective: Evaluate a repeated decrease model.

A decrease of r each period retains fraction 1−r. This factor lies between zero and one for ordinary partial decay. Applying the percentage to the original amount every period would instead produce linear change. Common error: subtracting the same original percent amount every period. Restrictions and interpretation: A growth factor is 1+r and a decay factor is 1−r; use a consistent time unit.

Formulas

$A(t)=A₀(1−r)^t, 0<r<1$

Worked examples

Try itA sample starts at 1472 grams and loses 25% of its mass per day. What mass is modeled after 2 days?Show answer

Each day retains 0.75 of the prior mass. Two days give 1472(0.75)²=1472(9/16)=828 grams. Therefore the requested answer is 828.

Solve for an initial amount

Objective: Recover the start from a later exponential value.

To reverse repeated growth or decay, divide the later amount by the complete factor raised to the elapsed-period exponent. Dividing by the rate itself does not undo the full model. Common error: dividing by a single factor when several periods have passed. Restrictions and interpretation: A growth factor is 1+r and a decay factor is 1−r; use a consistent time unit. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$A₀=A(t)/b^t, b>0$

Worked examples

Try itA quantity doubles every 3 days and is 2000 after 6 days. What was its initial value?Show answer

Six days contains 6/3=2 doubling periods. The multiplier is 2²=4. Divide 2000 by 4 to get 500. Therefore the requested answer is 500.

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