Exponentials, logarithms and modelling · Exponentielles, logarithmes et modélisation
| English | Français |
|---|---|
| half-life/hɑːf laɪf/ | demi-vie |
Why does a decay model stay positive?
- A medicine concentration falls by the same percentage each hour. A constant subtraction would eventually predict a negative amount.
- This lesson studies half-life 半衰期: The time for a decaying quantity to fall to half its initial value.
Choose the mathematical structure
- For y=Ae^(kt), k is a proportional rate. Taking logs gives ln y=ln A+kt. Logarithms require positive arguments, and log(x+y) is not log x+log y.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines half-life? · Quelle description définit correctement la demi-vie ?
The time for a decaying quantity to fall to half its initial value. · Le temps nécessaire pour qu'une quantité décroissante diminue de moitié par rapport à sa valeur initiale.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
If y=80e^(-0.2t), y=40 gives e^(-0.2t)=1/2. Hence t=ln2/0.2≈3.466. For 3^x=20, x=ln20/ln3. Plotting ln y against t linearises this exponential model.
Exponentials, logarithms and modelling · Exponentielles, logarithmes et modélisation
For y=Ae^(kt), k is a proportional rate · Pour y=Ae^(kt), k est un taux proportionnel
Explain why the half-life depends on 0.2 and not on the initial amount 80. · Expliquez pourquoi la demi-vie dépend de 0.2 et non de la quantité initiale 80.
Solve 2^x=32. · Résoudre 2^x=32.
32=2⁵, so x=5. · 32=2^⁵, donc x=5.
Test a tempting shortcut
- A fitted exponential is a model, not a guarantee. Specify the time units and range of use. A negative k describes decay; a negative starting amount usually contradicts the context.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
For positive x and y, ln(x+y)=ln x+ln y. This claim is false. Explain which definition or assumption it violates.
Find ln(e³). · Trouver ln(e³).
The natural logarithm undoes exponentiation: ln(e³)=3. · Le logarithme naturel annule l'exponentiation : ln(e³)=3.
For positive x and y, ln(x+y)=ln x+ln y. · Pour x et y positifs, ln(x+y)=ln x+ln y.
A fitted exponential is a model, not a guarantee. Specify the time units and range of use. A negative k describes decay; a negative starting amount usually contradicts the context. · Un ajustement exponentiel est un modèle, pas une garantie. Spécifier les unités de temps et la plage d'utilisation. Un k négatif décrit une décroissance ; un montant initial négatif contredit généralement le contexte.
Interpret a new situation
- Compare actual observations with the model. Systematic departures may indicate changing conditions. In a report, explain what the rate and initial value mean, rather than giving a bare equation.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the half-life of 80e^(-0.2t), to three decimal places. · Trouver la demi-vie de 80e^(-0.2t), à trois décimales près.
Set e^(-0.2t)=1/2. Then t=ln(2)/0.2≈3.466 to three decimal places. · Poser e^(-0.2t)=1/2. Alors t=ln(2)/0.2≈3.466 à trois décimales près.
Match each part of a complete solution to its purpose. · Associer chaque partie d'une solution complète à son but.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Une hypothèse justifie le modèle ; une vérification teste le résultat ; l'interprétation le relie à la question.
Use this in your course
- Current first-assessment-2021 Analysis and Approaches SL. This is authored concept support; the full guide is needed to certify every objective.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The time for a decaying quantity to fall to half its initial value. Choose the relationship, show the method, check its assumptions and interpret the result.