Exponential Growth and Decay by Percentage

Both growth and decay by percentage come from the same basic formula:

\( A(t) = P(1 \pm r)^t \)

• \(P\) is the initial (starting) amount, when \(t = 0\).
• \(r\) is the percentage rate, written as a decimal (so 5% becomes \(0.05\)).
• \(t\) is the number of time periods that have passed.
• \(A(t)\) is the amount after \(t\) periods.

The sign in front of \(r\) tells you whether you have growth or decay:

• Growth: \( A(t) = P(1 + r)^t \), where the base \(1 + r\) is greater than \(1\).
• Decay: \( A(t) = P(1 - r)^t \), where the base \(1 - r\) is between \(0\) and \(1\).

Notice that both versions are exponential functions with \(t\) in the exponent, so their graphs share the same overall shape you studied when you learned to graph exponential functions: one curve rises without bound, and the other curve flattens out and approaches (but never reaches) zero.

A town has a population of \(1000\) people, and the population grows by \(5\%\) each year. Write a function for the population after \(t\) years, and find the population after \(10\) years.

Step 1: Identify \(P\) and \(r\). Here \(P = 1000\) and \(r = 0.05\), and since the population is increasing, we use the growth formula.

Step 2: Write the function.

\( A(t) = 1000(1 + 0.05)^t = 1000(1.05)^t \)

Step 3: Substitute \(t = 10\).

\( A(10) = 1000(1.05)^{10} \approx 1628.9 \)

So after \(10\) years, the population is about \(1629\) people. The graph below shows how this population grows over \(20\) years, starting from \(1000\) and curving upward faster and faster.

A car is purchased for \($20{,}000\) and depreciates (loses value) by \(8\%\) each year. Write a function for its value after \(t\) years, and find its value after \(6\) years.

Step 1: Identify \(P\) and \(r\). Here \(P = 20000\) and \(r = 0.08\), and since the value is decreasing, we use the decay formula.

Step 2: Write the function.

\( A(t) = 20000(1 - 0.08)^t = 20000(0.92)^t \)

Step 3: Substitute \(t = 6\).

\( A(6) = 20000(0.92)^6 \approx 12240.6 \)

The car is worth about \($12{,}241\) after \(6\) years. Notice how the value keeps dropping toward zero but never actually reaches it, which is the hallmark of exponential decay.

Before writing any equation, decide whether the quantity is increasing or decreasing over time:

• Words like "increases," "grows," "appreciates," or "doubles" point to growth, so you add \(r\): base \( = 1 + r\).
• Words like "decreases," "decays," "depreciates," "shrinks," or "loses value" point to decay, so you subtract \(r\): base \( = 1 - r\).

A special and very common case of percentage decay is when the rate is exactly \(50\%\) per fixed period. That situation is called half-life, and it has its own shortcuts worth knowing; see exponential decay and half-life for that specific version of the decay formula.

• Forgetting to convert the percentage to a decimal. A rate of \(7\%\) means \(r = 0.07\), not \(r = 7\).
• Mixing up the sign. Always check the wording carefully: growth uses \(1 + r\), decay uses \(1 - r\).
• Using the wrong exponent. The exponent \(t\) must match the same time unit as the percentage rate (for example, if the rate is per year, \(t\) must be in years).

Percentage-based growth and decay problems set up the equations you will later need to solve for time, which requires isolating an exponent using logarithms; that skill is covered in solving logarithmic equations. Being comfortable writing \(A(t) = P(1 \pm r)^t\) correctly is the foundation for all of that later work.

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