Math · Functions

Exponential Change / Growth, Decay, and Logarithms

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Derivative as Slope / Local Rate of ChangeCarry changing steepness into local slope on the graph

Key takeaway

The sign of k turns the same exponential model into growth or decay.
Equal time steps multiply the amount, which creates fixed doubling-time or half-life cadences.
Target-time questions are logarithmic because they ask when the exponential curve reaches a chosen ratio.
A graph crossing, a log-view crossing, and a doubling/half-life readout should tell the same story.

Common misconception

Do not treat logarithms as a detached new topic here. The logarithm is the inverse-time question for the same exponential curve.

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

Keep this idea moving

Open the next concept, route, or track only when you want the current model to widen into a larger branch.

Carry changing steepness into local slope on the graphCalculus

Derivative as Slope / Local Rate of Change

Slide a point along one curve, tighten a secant into a tangent, and connect local steepness to the derivative graph without leaving the same live bench.

See the same decay law inside a physics bench with real half-life languageModern Physics

Radioactivity and Half-Life

Use one compact decay bench to see why each nucleus decays unpredictably, why large samples still follow a regular half-life curve, and how to read remaining-count graphs honestly.

Next in Rational functions and asymptotesFunctions

Rational Functions / Asymptotes and Behavior

Vary one shifted reciprocal family so domain breaks, vertical and horizontal asymptotes, intercepts, and removable-hole behavior stay tied to the same graph.

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Reference and support

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Equation snapshotExponential model · Solve for the target-hit timeShow 3 equations

Read the model first as a live curve, then use the log equation only when the question is turned around to solve for time.

Exponential model
$y (t) = y_{0} e^{k t}$
Uses one starting value y_0 and one continuous rate k to model growth or decay.
Solve for the target-hit time
$t_{*} = \frac{l n ( T / y _{0} )}{k}$
Turns the target ratio into a logarithm because time appears in the exponent.
Fixed exponential time scales
$t_{double} = \frac{l n 2}{k}, T_{1/2} = \frac{l n 2}{∣ k ∣}$
Shows that doubling time for growth and half-life for decay depend on the rate, not on the current amount.

Why it behaves this way

Explanation

Exponential change is easier to understand when growth or decay stays attached to one live curve. This bench keeps the starting value, the continuous rate, the target line, and the target-hit time on the same graph so every algebra step matches a visible crossing.

The key idea is multiplicative change, not adding the same amount each step. Equal time steps multiply the amount by the same factor, which is why growth can double and decay can halve on a fixed schedule. The logarithm appears when you reverse the question and solve for the time needed to reach a chosen target.

Key ideas

01In an exponential model, the amount follows y(t) = y_0 e^{kt}, and 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 a constant amount each step.

03The inverse question is logarithmic: compare the target to the start with T / y_0, then solve for time with ln(T / y_0) / k.

Worked examples

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Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the growth-to-target preset and keep the target marker and log guide visible. First compare the target to the start, then solve for the hit time, then check that time against the graph.

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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 the curve take to hit the target?

y_{0}

Initial value

k

Rate

0.25 1/time

T

Target

T / y_{0}

Target ratio

1. Compare the target to the start

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

2. Solve for time with the log rule

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

3. Check the answer against doubling time

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.

Quick test

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Growth to target preset

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Featured setups

Growth to a target

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

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 6

Functions and Change

Exponential Change / Growth, Decay, and Logarithms appears later in this track, so it is cleaner to start from the beginning first.

Previous step: Rational Functions / Asymptotes and Behavior

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