Exponential change rule
Uses one starting value and one continuous rate to generate exponential growth or decay.
Initial value 3 Rate 0.25 1/time
Concept module
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.
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. The exponential curve grows from 3 with continuous rate k = 0.25. The target 12 is reached at about t = 5.55. The doubling time is about 2.77, where the amount reaches 6. The inverse question becomes logarithmic because ln(target / y0) = 1.39.
Interactive lab
Exponential change
Keep the live curve, the opposite-rate comparison, and the logarithmic target question on one bench so growth, decay, and inverse time stay visually tied together.
Drag the target line or use the sliders
Controls
Set the starting height of the exponential curve.
Use positive values for growth and negative values for decay.
Choose the amount you want the curve to reach.
More tools
Secondary controls, alternate presets, and less-used toggles stay nearby without crowding the main bench.
Show
More tools
Secondary controls, alternate presets, and less-used toggles stay nearby without crowding the main bench.
More presets
Presets
Predict -> manipulate -> observe
Keep the active prompt next to the controls so each change has an immediate visible consequence.
ComparePrompt 1 of 4
Switch the sign of the rate and keep the same start value. One curve rises, the other falls, and the target only stays reachable on one side of the start.
Graphs
Switch graph views without breaking the live stage and time link.
Amount vs time
Shows the current exponential curve, the opposite-rate comparison, and the target marker on the same amount-versus-time graph.
time: 0 to 8amount: 0 to 32
Current growthOpposite-rate comparisonTargetCurrent target hit
Equation map
See each variable before you move it.
Select a symbol to highlight the matching control and the graph or overlay it most directly changes.
Initial value
3Sets the starting height of both the current curve and the opposite-rate comparison curve.
Graph: Amount vs timeGraph: Log target viewOverlay: Target markerOverlay: Log guide
Equations in play
Choose an equation to sync the active symbol, control highlight, and related graph mapping.
More tools
Detailed noticing prompts, guided overlays, and challenge tasks stay available without taking over the main bench.
Hide
More tools
Detailed noticing prompts, guided overlays, and challenge tasks stay available without taking over the main bench.
What to notice
Use these prompts when the target-time algebra starts to drift away from the curve.
ComparePrompt 1 of 4
Graph: Amount vs time
Switch the sign of the rate and keep the same start value. One curve rises, the other falls, and the target only stays reachable on one side of the start.
Control: RateControl: TargetGraph: Amount vs timeOverlay: Opposite-rate comparisonOverlay: Target markerEquationEquation
Guided overlays
Focus one overlay at a time to see what it represents and what to notice in the live motion.
4 visible
Overlay focus
Target marker
Show the target line and any reachable hit point on the main curve.
What to notice
- The hit point only appears when the target sits on the side of the start that matches the sign of the rate.
Why it matters
It keeps the inverse question tied to one visible crossing instead of a detached equation.
Control: TargetControl: RateGraph: Amount vs timeEquationEquation
Challenge mode
Use the curve, target line, and half-life cue together. The goal is to read a decay target as a multiplicative story, not just to land one number by accident.
0/1 solved
TargetCore
4 of 7 checks
Quarter-target checkpoint
Build a decay case where the target is about one quarter of the start, so the curve reaches it in about two half-lives.
Graph-linkedGuided start2 hints
Suggested start
Start from a simple decay case, then lower the target and sharpen the decay until the target-time readout matches two half-life steps.
Matched
Open the Amount vs time graph.
Amount vs timeMatched
Keep the Target marker visible.
OnMatched
Keep the Doubling or half-life guide visible.
OnMatched
Keep the Log guide visible.
OnPending
Keep the start value near , between and .
3Pending
Keep the continuous decay rate between and .
0.25Pending
Set the target near one quarter of the start, between and .
12The checklist updates from the live simulation state, active graph, overlays, inspect time, and compare setup.
The exponential curve grows from 3 with continuous rate k = 0.25. The target 12 is reached at about t = 5.55. The doubling time is about 2.77, where the amount reaches 6. The inverse question becomes logarithmic because ln(target / y0) = 1.39.
Equation detailsDeeper interpretation, notes, and worked variable context.
Target time from a logarithm
Solves the inverse question by turning the multiplicative target ratio into a logarithm.
Initial value 3 Rate 0.25 1/time Target 12
Fixed exponential time scales
Shows the fixed time scale for doubling in growth and halving in decay.
Rate 0.25 1/time
Progress
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Let the live model runChange one real controlOpen What to notice
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Stable links
Starter track
Step 3 of 60 / 6 completeFunctions 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.
Short explanation
What the system is doing
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.
Worked example
Read the full frozen walkthrough.
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.
Live worked examples are available on Premium. You can still read the full frozen walkthrough on the free tier.
View plansFrozen 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.
Prediction prompt
Decide whether the target should sit just below the start or much lower before you test it.
Check your reasoning
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?
Choose one answer to reveal feedback, then test the idea in the live system if a guided example is available.
Accessible description
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.
Read next
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These suggestions come from the concept registry, so the reason label reflects either curated guidance or the fallback progression logic.
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