Exponential Growth and Decay: Doubling Time, Half-Life, and Graphs

Adding versus multiplying

A savings jar that gains $50 a month grows linearly: the same amount is added each step. An account that gains 5% a year grows exponentially: each year's increase is bigger than the last, because 5% of a bigger balance is more money.

The quick test is to look at a table of values. If the differences between steps are constant, the pattern is linear. If the ratios are constant, it is exponential.

The basic model

y = a · b^t. Here a is the value when t = 0 and b is the factor applied every time step.

Growth by a rate r per step means b = 1 + r, so 6% growth is b = 1.06. Decay by a rate r means b = 1 - r, so losing 20% each step is b = 0.80.

When the growth is stated as a doubling or halving period, put the period in the exponent. A culture of 500 bacteria that doubles every 3 hours is N = 500 · 2^(t/3). After 12 hours that is 500 · 2^4 = 8000.

Half-life

Decay is often described by its half-life, the time it takes to lose half of what is there. The model is A = A0 · (1/2)^(t/h), where h is the half-life.

A dose of 80 mg with a half-life of 6 hours leaves 80 · (1/2)^3 = 10 mg after 18 hours: 40 after the first 6 hours, 20 after 12, and 10 after 18.

Doubling time and the rule of 72

To find how long growth at rate r takes to double, solve (1 + r)^t = 2. Taking logs gives t = ln(2) / ln(1 + r). At 7% per year that is ln(2) / ln(1.07) ≈ 10.24 years.

The rule of 72 is a mental shortcut: divide 72 by the percentage rate. 72 / 7 ≈ 10.3 years, close to the exact answer. It works best for rates between about 4% and 12%.

Compound and continuous interest

Money is the most common exponential model. Compounded n times a year, A = P(1 + r/n)^(nt). Compounded continuously, A = Pe^(rt).

$1,000 at 6% for 12 years grows to about $2,012.20 compounded yearly and about $2,054.43 compounded continuously. More frequent compounding always helps, but the gain shrinks as n grows.

For continuous growth the doubling time is exactly ln(2) / r. At 5% that is about 13.86 years.

Graphing y = a·b^(x - h) + k

The parent graph y = b^x passes through (0, 1) and has a horizontal asymptote at y = 0. Each letter moves or reshapes it.

h shifts the graph right by h. k shifts it up by k, which also moves the asymptote to y = k. a stretches the graph vertically, and a negative a reflects it over the asymptote.

The domain is all real numbers. The range is y > k when a is positive and y < k when a is negative.

Example: y = 2 · 3^(x - 1) - 4 has its asymptote at y = -4 and range y > -4. It passes through (1, -2), because 3^0 = 1, and (2, 2), because 2 · 3 - 4 = 2.

Solving for time

Questions such as 'when will the population reach 10,000?' need a logarithm. Isolate the power and take logs of both sides: 500 · 2^(t/3) = 10000 gives 2^(t/3) = 20, so t/3 = ln(20) / ln(2) ≈ 4.32 and t ≈ 12.97 hours.

Calculator programs for this topic

These free TI-84 Plus CE Python programs carry out the steps above. Each page explains its own math, shows a screen preview, and lets you read the source before you download it.

• DuckieDai Compound Interest: Finds the final amount, starting amount, rate comparison, or time for compound and continuous interest.
• DuckieDai Exponential Transform Solver: Reads y = a·b^(x-h)+k: growth or decay, shifts, asymptote, domain, range, and points to plot.
• DuckieDai Log and Exponent Solver: Solves b^x = y for the exponent, which is how you find a doubling time or a time to reach a target.