Exponential and Log Models

Many real-world quantities change by the same factor in every equal step of time: a population multiplies, savings earn interest, a radioactive sample halves. Each is an exponential model:

y = A \cdot k^{\,t}

Here A is the starting amount (the value when t = 0), k is the growth or decay factor per unit of time, and t is the time. This is the same multiply-by-a-constant pattern you have seen in , now read off a smooth curve. You will use logs to solve for the exponent when the unknown is the time.

Reading A and k

Two numbers tell the whole story. Set t = 0: since k^0 = 1, the value is A — the starting amount. Then each extra unit of time multiplies by k:

y(0) = A, \qquad y(1) = A\,k, \qquad y(2) = A\,k^2.

The sign of "growth versus decay" lives entirely in k:

Growth when k > 1 — the value rises (a doubling population has k = 2).
Decay when 0 < k < 1 — the value falls (a sample that halves has k = \tfrac{1}{2}).

A percentage rate folds neatly into k: 5% yearly interest is k = 1.05; losing 5% a year is k = 0.95. These are the seen as a formula.

See it: growth versus decay

Slide A to set where the curve starts, and slide k across 1. Above 1 the curve climbs (growth); below 1 it sinks toward zero (decay); at exactly 1 it is flat. Time only runs forward, so t \ge 0.

Using logs to find the time

When the unknown is when a model reaches a target value, the time t sits in the exponent — so take a logarithm. To find when a colony of A = 200 bacteria that doubles each hour (k = 2) reaches 1600:

200 \cdot 2^{\,t} = 1600 \;\Longrightarrow\; 2^{\,t} = 8 \;\Longrightarrow\; t = \log_2 8 = 3 \text{ hours.}

Divide by the starting amount, then log both sides to bring the exponent down. The same recipe finds a half-life, a doubling time, or how long savings take to reach a goal.

Khan Academy works through exponential model word problems here: