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Exponential Change / Growth, Decay, and Logarithms

Change one starting value, one rate, and one target so growth, decay, doubling or half-life, and logarithmic target time all stay tied to the same live curve.

Interactive lab

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Featured setups

Growth to a target

Open a clean growth case with the target marker, opposite-rate curve, and log guide all visible.

Decay and half-life

Start from a decay case and keep the half-life guide visible while the target sits below the start.

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Starter track

Step 3 of 60 / 6 complete

Functions and Change

Earlier steps still set up Exponential Change / Growth, Decay, and Logarithms.

1. Graph Transformations2. Rational Functions / Asymptotes and Behavior3. Exponential Change / Growth, Decay, and Logarithms4. Derivative as Slope / Local Rate of Change+2 more steps

Previous step: Rational Functions / Asymptotes and Behavior.

Start from track beginning

Why it behaves this way

Explanation

Exponential change is easiest to trust when the growth or decay stays attached to one live curve instead of a detached rule sheet. This bench keeps the starting value, the continuous rate, the target line, and the inverse-time question tied to the same graph.

The key move is multiplicative rather than additive. Equal steps in time multiply the amount by the same factor, which is why growth can double and decay can halve on a fixed schedule. The logarithm appears only when you turn the question around and ask how long it takes to reach a chosen target.

Key ideas

01In an exponential model, the amount follows y(t) = y_0 e^{kt}, so the sign of k decides whether the curve grows or decays.
02Doubling time and half-life are fixed time scales because exponential change repeats by multiplication, not by adding the same amount each step.
03The inverse question is logarithmic: solving for time means taking ln(target / y_0) and dividing by the rate.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough
This concept keeps the bench compact, so the main curve, the target line, the cadence readout, and the log view carry the core reasoning directly.

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Frozen valuesUsing frozen parameters

A population starts at 3 units and grows at a continuous rate of 0.25 until it reaches 12 units. How long should that take?

Initial value

3

Rate

0.25 1/time

Target

12

Target ratio

4

1. Compare the target to the start

The target is four times the starting value, so \(T / y_0 = 12 / 3 = 4\).

2. Use the inverse-time rule

Solve \(t_* = \ln(T / y_0) / k\), which gives \(t_* = \ln 4 / 0.25 \approx 5.55\).

3. Check the graph against the cadence

Because \(\ln 4 = 2 \ln 2\), this target sits at exactly two doubling times, so the hit marker should land a little past 5.5 time units.

Target hit time

The curve reaches the target after about 5.55 time units.
This keeps the logarithm tied to one visible crossing: the growth curve must double twice to turn 3 into 12.

Common misconception

Exponential change means the graph always rises quickly, and logarithms are a separate chapter with no direct connection to the curve.

A negative rate gives honest exponential decay, so the same model can fall toward zero instead of rising away from the start.

The logarithm is what appears when you solve the exponential target equation for time. It is the inverse question, not a disconnected new object.

Mini challenge

Set a decay case where the target is below the start and the target time is longer than one half-life.

Make a prediction before you reveal the next step.

Decide whether the target should sit just below the start or much lower before you test it.

Check your reasoning against the live bench.

The target must sit well below the starting value so the curve needs more than one half-life before it reaches it.
One half-life only cuts the amount in half. If the target is much lower than that first halfway mark, the inverse-time readout has to land later than one half-life.

Quick test

Variable effect

Question 1 of 3

Answer from the live curve and the inverse-time readout together.

Which single change turns the same exponential bench from growth into decay?

Use the live bench to test the result before moving on.

Accessibility

The simulation shows an exponential amount-versus-time curve with a starting value, a target line, an opposite-rate comparison curve, a doubling-time or half-life guide that lands on the matching amount, and a smaller log view that straightens the inverse target question.

Graph summary

One graph shows the current exponential curve, the opposite-rate comparison, the target crossing, and the one-step doubling or half-life amount when that cue exists. A second graph shows ln(amount / initial) as a straight line with the matching target-log line.