Your child can solve a logarithm equation but struggles to turn “grows by a percentage” or “falls to half” into a useful model. Secondary 4 Additional Mathematics tuition should begin with what changes over equal time intervals. An exponential model multiplies by a fixed factor; a linear model adds a fixed amount. Naming that distinction makes the formula meaningful before any calculator work begins.
Secondary 4 Additional Mathematics tutorials can build a clear route from initial value and time unit to factor, unknown parameter and requested threshold. The student should label each quantity, keep a ratio separate from a difference and check the answer in the original model. A correct logarithm calculation cannot rescue an incorrectly chosen growth factor.
A Secondary 4 Additional Mathematics tutor can inspect the model line beside the question’s wording. Parents can bring the original data, percentage interpretation, time units and first independent attempt. Repair the missing modelling decision, then change the context or time interval to test whether the understanding transfers.
eduKateSG · Secondary 4 Additional Mathematics
Find the question closest to your family
Choose a route, read the explanation, and use only the worked checks that fit your child’s current course.
ROUTE 1 · CHAPTERS 1–2
Understand the concern
Equal intervals mean equal factors in an exponential model
ROUTE 4 · CHAPTERS 10–15
See what the work reveals
Bring the model line, not only the calculator answer
Full chapter index · Worked learning checks · Additional Mathematics tuition guide
| What the work shows | First teaching response | Later independent check |
|---|---|---|
| First interval already wrong | Translate percentage to retained factor | Fresh increase versus decrease |
| Rate calculation mismatches data | Check elapsed-time ratio and units | Verify both observations |
| Threshold time plausible, output wrong | Inspect baseline and permitted time steps | Original-model threshold comparison |
CHAPTER 1 OF 17 · Understand the concern
1. Equal intervals mean equal factors in an exponential model
A discrete exponential model can be written Q(t)=Abᵗ when one unit of t matches the stated interval. A is the initial quantity at t=0, and b is the factor applied per time unit. Repeated multiplication produces the power bᵗ.
A constant percentage growth corresponds to a constant multiplicative factor greater than one. A constant percentage decline corresponds to a factor between zero and one. This differs from adding or subtracting the same amount each interval.
For example, a quantity starting at 100 and multiplying by 1.2 gives 100,120,144 across consecutive intervals. Adding 20 each time would instead give 100,120,140. The first step matches, but the second reveals different models.
Read whether the problem states a model, gives exact mathematical data or asks the student to infer an appropriate relationship. Real observations do not automatically establish indefinite exponential behaviour. The examples here are mathematical illustrations under their stated assumptions.
Parents can ask, “What stays the same each interval: a difference or a ratio?” That question helps the tutor identify model choice before judging the later algebra.
CHAPTER 2 OF 17 · Understand the concern
2. A percentage change is not the multiplier itself
A growth of p percent per interval gives factor 1+p/100. A decline of p percent gives factor 1−p/100, for a meaningful nonnegative retained fraction. The new quantity includes the retained old amount as well as the change.
A 20 percent increase therefore uses 1.2, not 0.2. A 20 percent decrease uses 0.8, not −0.2. The sign of a decline is represented by a factor below one; the modelled positive quantity need not become negative.
Distinguish “falls by 20 percent” from “falls to 20 percent.” The first retains 80 percent; the second retains 20 percent. Those words create very different multipliers and should be read before calculation.
If the stated interval is three hours, a factor of 0.8 belongs to each three-hour interval. The exponent for elapsed time t hours is t/3 in the corresponding mathematical model, not t without changing the time unit.
The tutor can use one short step to check percentage interpretation before introducing logarithms. If the first interval quantity is wrong, repair the factor rather than ask the student to practise longer threshold equations.
The exponent must count the model’s intervals consistently. If a quantity multiplies by b every h hours, Q(t)=Abᵗ⁄ʰ with t measured in hours is a natural form. At t=h, the exponent is one and the factor is applied once.
Changing the time unit changes the numerical rate parameter in an equivalent expression. The quantity described should remain the same. Do not copy a per-hour parameter into a per-minute formula without conversion.
In a continuous form Q(t)=Aeᵏᵗ, the product kt must be dimensionless. If t is in hours, k has a per-hour interpretation; if t is in days, its numerical value changes for the equivalent model.
A negative k gives decay and a positive k gives growth when A>0. The initial value remains A because e⁰=1. A shift in time origin changes how the initial condition is represented, so label when t=0 occurs.
Parents can ask which elapsed time makes the exponent equal one. That simple check exposes a hidden unit mismatch before it spreads into parameter recovery or a predicted threshold.
CHAPTER 4 OF 17 · Plan the support
4. Use the initial condition before solving parameters
At t=0, a model Q(t)=Abᵗ gives Q(0)=A. The same is true of Aeᵏᵗ. An initial value can therefore determine A directly when the time origin matches the stated start.
If the first observation occurs at another time, it does not automatically equal A. Use the model at that time to form an equation. Two suitable observations may determine A and the growth factor or rate through their ratio and substitution.
Taking a ratio can eliminate A: Q(t₂)/Q(t₁)=bᵗ²⁻ᵗ¹ or eᵏ⁽ᵗ²⁻ᵗ¹⁾. The elapsed difference matters. Do not substitute t₂ alone if the first observation was not at zero.
After solving parameters, return to both observations. For exact mathematical data they should agree exactly or to the stated numerical precision. For measured data a fitted model may be approximate, and that should be acknowledged as part of the task.
The tutor can inspect the parameter equations before the final values. A wrong time difference or observation pairing is a modelling fault, while an incorrect logarithm operation after a correct ratio is a different algebraic boundary.
When the quantity is specified and time is unknown, the time appears in an exponent. Isolate the exponential factor first, then apply a valid logarithm to solve for the exponent. Keep the initial value and any offset in their proper roles.
For Abᵗ=C with A>0, b>0 and b≠1, divide by A to obtain bᵗ=C/A. With a positive target ratio, t=ln(C/A)/ln b. Any consistent valid logarithm base gives the same ratio.
In Aeᵏᵗ=C with k≠0, the form is t=ln(C/A)/k. A target and a model domain may restrict whether the resulting time is relevant. A negative algebraic time can lie outside a problem that begins at t=0.
Do not take logarithms of a sum as if it were a product. If the model has a baseline L plus an exponential term, subtract L before isolating the factor. The inverse operation must match the actual expression.
The independent task should change the initial value, factor or target. A student who merely inserts remembered numbers into a threshold formula has not shown that the model can be interpreted afresh.
CHAPTER 6 OF 17 · Build a workable learning loop
6. Half-life and doubling time describe ratios
A half-life is the time over which the exponential quantity or relevant excess halves under the model. A doubling time is the time over which it doubles. These are ratio conditions, not fixed amounts removed or added.
For Q(t)=Aeᵏᵗ with k<0, the half-life h satisfies eᵏʰ=1/2, so h=ln(1/2)/k, a positive value. For k>0, the doubling time d satisfies eᵏᵈ=2, giving d=ln 2/k.
The same half-life repeats for a simple exponential decay because equal elapsed intervals multiply by the same factor. It takes the same duration to go from A to A/2 as from A/2 to A/4. The absolute decrease differs.
If the model approaches a nonzero baseline, clarify which quantity halves: the excess above the baseline, not necessarily the full measured quantity. The wording and model determine the ratio to use.
The tutor should connect the named time to the exponential factor before using a memorised expression. That makes the method usable even when the question describes the ratio without using the term “half-life.”
CHAPTER 7 OF 17 · Build a workable learning loop
7. A baseline changes the quantity that decays
A model such as Q(t)=L+Ae⁻ᵏᵗ, with A>0 and k>0, approaches baseline L as t increases. The excess Q−L decays exponentially. The total quantity Q does not generally halve at the excess’s half-life.
At t=0, Q(0)=L+A. The coefficient A is the initial excess, not the full initial quantity. If the starting quantity and baseline are given, subtract to find A before solving the rate.
For a target C>L, isolate e⁻ᵏᵗ=(C−L)/A. A target equal to L is approached but never reached at finite real time by the positive exponential term. A target below L is impossible within this positive-excess model.
These conclusions are model interpretations, not universal statements about a real physical system. The given assumptions and allowed time range matter. Use the model only as the question or lesson specifies.
Parents can ask which part of the expression changes and which stays fixed. This highlights a common first fault: halving the full starting quantity when only the excess above an ambient or reference level follows the exponential factor.
CHAPTER 8 OF 17 · Build a workable learning loop
8. Continuous time and whole-step decisions are different outputs
A mathematical exponential model may allow real elapsed time, while a question asks for the first complete day, year or inspection step at which a threshold is met. Solve the continuous threshold first when appropriate, then interpret the allowed reporting times.
For a strictly increasing model, the first integer step reaching or exceeding a threshold is the smallest permitted integer not below the continuous crossing time. Check that integer and the previous one in the original model. A strict inequality needs special attention if equality occurs exactly at an integer.
For decay, the direction of the threshold comparison changes, but the same original-model checking habit helps. Do not round a crossing time to the nearest integer if the request is the first step satisfying a condition.
State the time unit and the start convention. Whether t=0 counts as an inspection point or a starting observation depends on the problem. The final answer should reflect the actual wording, not a universal rounding habit.
The tutor can separate logarithm solving from threshold interpretation. A correct decimal time followed by a wrong integer answer needs an output-decision repair, not another full lesson on logarithm laws.
CHAPTER 9 OF 17 · Build a workable learning loop
9. Check reasonableness and acknowledge model limits
A positive growth model with factor above one should increase over the allowed forward time. A positive decay model with factor between zero and one should decrease towards zero, or towards its specified baseline. A prediction with the opposite trend deserves inspection.
Use the initial condition, one known observation and a threshold check where relevant. These correspond to different aspects of the model. A matching initial value alone does not confirm the rate; a matching rate alone does not confirm the start quantity.
Keep exact expressions or sufficient calculator precision until final reporting. Early rounding of a rate can create visible discrepancies in a longer time prediction. Follow the actual requested accuracy without claiming a universal marking tolerance.
A mathematical model can be useful without being a permanent description of reality. Data range, assumptions and changing conditions matter in genuine applications. Label hypothetical examples clearly, and do not present a classroom calculation as a verified future outcome.
Parents can ask whether the proposed answer increases or decreases as expected and which original datum checks the rate. These are practical questions that help the tutor identify a wrong factor, sign or time unit.
CHAPTER 10 OF 17 · See what the work reveals
10. Bring the model line, not only the calculator answer
Bring the original wording, defined variables, time unit, model equation, initial condition, parameter equations and first independent threshold calculation. Include teacher feedback, school topic list, subject level and examination year. This evidence lets the tutor choose a focused starting point.
The official SEAB sources distinguish 2027 SEC G3 Additional Mathematics K341 and G2 K232 from 2026 O-Level Additional Mathematics 4049. Confirm the actual route with the school. The examples below are selected teaching tasks rather than a complete syllabus, assessment forecast or guaranteed outcome.
Ask which boundary needs repair: difference versus ratio, percentage factor, elapsed time, parameter recovery, baseline subtraction or final threshold reporting. The next fresh problem should change that decision meaningfully.
eduKateSG small-group tutorials use up to three students. Confirm current suitability, available times, fees, duration, location and attendance arrangements directly. Agree on manageable continuation work that the student can attempt honestly before the next review.
The aim is an understandable model with a verified route back to the requested quantity. A student who can explain the factor and time unit has a stronger basis for using logarithms than one who reaches a decimal without knowing what its exponent counted.
CHAPTER 11 OF 17 · See what the work reveals
11. A percentage increase followed by the same decrease does not cancel
A hypothetical quantity starts at 100, increases by 10 percent and then decreases by 10 percent of its new value. The first factor is 1.1 and the second is 0.9. The final quantity is 100×1.1×0.9=99, not 100.
The percentage changes apply to different current quantities. The increase is 10, while the later decrease is 11. Equal stated percentages therefore do not mean equal absolute amounts or inverse factors.
To reverse a factor of 1.1 exactly, multiply by 1/1.1=10/11. From 110, that returns 100. The required decrease fraction is 1−10/11=1/11, about 9.0909 percent of the increased quantity. This is an inverse-factor calculation, not a rule to negate the original percentage.
For a fresh 20 percent increase followed by a 20 percent decrease, a starting 200 becomes 200×1.2×0.8=192. The factors multiply to 0.96. The same structural explanation holds with different amounts.
If this paired change repeats each cycle under a stated model, the retained factor per complete cycle is 0.99 in the first example. After two cycles from 100, the quantity is 98.01. The time variable must count complete cycles if that factor is used directly.
Ask the learner what quantity each percentage refers to before constructing the formula. A student who can calculate powers accurately may still treat equal percentage labels as cancelling automatically. The tutor can repair the factor interpretation with this short contrast and then return to the actual exponential task. Keep the example hypothetical and tied to its stated assumptions; it is a mathematical illustration rather than a prediction about a real asset or process.
CHAPTER 12 OF 17 · See what the work reveals
12. An imagined modelling review: the rate was calculated for the wrong quantity
Imagine a student given the hypothetical model T(t)=18+72e⁻ᵏᵗ, with k>0 and t measured in hours. The quantity begins at T(0)=90 and has T(6)=54. The student sees that 54 is 60 percent of 90 and uses that ratio to find the decay rate. The logarithm operation may be performed accurately, but the changing quantity has been misidentified.
This is a teaching illustration, not an observed real process or learner result. The model's constant baseline is 18. The initial excess is 72, while the six-hour excess is 54−18=36. It is the excess that halves. The correct equation is 36=72e⁻⁶ᵏ, giving e⁻⁶ᵏ=1/2 and k=ln 2/6.
Preserve any secure inverse-exponential work the learner showed. The first repair is baseline subtraction before the ratio. Label total T, baseline 18 and excess T−18. Ask which of these appears as the exponential multiple. This makes the model line, rather than a memorised half-life formula, the source of the decision.
At twelve hours the excess has halved twice, so T(12)=18+72/4=36. The total sequence is 90,54,36 at six-hour steps, while the excess sequence is 72,36,18. The total ratios are not all one half. Comparing both sequences shows why halving the total would describe a different model.
Suppose the next request is the real time at which T=27. Subtract the baseline: 27−18=9. Then e⁻ᵏᵗ=9/72=1/8. This requires three excess half-lives, so t=18 hours. The logarithm route gives the same answer: −kt=ln(1/8)=−3ln 2, and k=ln 2/6 yields t=18.
Check the original model at eighteen hours. The contribution is 72×1/8=9, and restoring the baseline gives 27. The original initial and six-hour observations also remain valid. These are distinct checks: start quantity, rate and requested threshold.
Now ask whether the model reaches T=18 at a finite time. That would require e⁻ᵏᵗ=0, which is impossible for finite real t. The model approaches 18 as forward time increases. A target below 18 is also impossible in this positive-excess model. Do not present the asymptote as a finite arrival merely because a displayed contribution becomes very small.
For an independent baseline variant, use Q(t)=12+48e⁻ᵏᵗ with Q(4)=36. The initial total is 60, the four-hour excess is 24, and k=ln 2/4. At eight hours Q=24. To reach Q=18, the excess must fall from 48 to 6, three halvings, so t=12. Let the learner form the excess ratio with the earlier worked solution closed.
A separate percentage-factor variant can use a model with no baseline: Q(n)=100(0.8)ⁿ at whole steps. The first step at which Q<50 is n=4, because Q(3)=51.2 and Q(4)=40.96. This tests discrete threshold reporting after model selection. It should not be confused with the earlier baseline-subtraction repair.
The real crossing time for that last model is ln(0.5)/ln(0.8), about 3.1063. Rounding to the nearest whole step would give 3, but step 3 fails the original strict threshold. The neighbouring-value check explains why the final answer is 4. If the learner's logarithm time is correct, preserve it while teaching the whole-step decision.
The tutor can therefore identify different secure and insecure boundaries: percentage interpretation, time interval, baseline subtraction, inverse solving and permitted output steps. A mistake in one does not prove that all exponential work is weak. Conversely, a correct decimal after supplied modelling hints does not establish independent model construction.
Parents can ask which quantity is multiplied by the factor and which original observation checks the recovered rate. Bring the actual wording and model line, including any baseline and time domain. Those details let the tutor choose a suitable first task and an honest fresh continuation.
This sequence does not claim that a real quantity follows the model indefinitely or that one lesson guarantees a particular result. It shows how a learner can make the route understandable and verifiable. Return to the actual school question once the smaller boundary is repaired, preserving the model's assumptions and the requested reporting accuracy.
CHAPTER 13 OF 17 · See what the work reveals
13. Worked learning checks: first decisions
1. A percentage growth factor
Try first. A hypothetical quantity starts at 200 and grows 10 percent per day under a model. Find its day-three value.
Worked reasoning. Use Q(t)=200(1.1)ᵗ with t in days. At t=3, Q=200×1.331=266.2. The factor 1.1 includes the retained whole and the additional 10 percent. It applies per day, so three days correspond to exponent three.
Check. Consecutive values are 200,220,242,266.2. Their ratios are 1.1 while their differences are not constant.
Error to notice. Using 0.1ᵗ would describe retaining only a tenth each day, not growing by 10 percent.
Independent variant and answer. Starting at 300 with 20 percent growth per interval gives 518.4 after three intervals.
What this tells the tutor. Ask what the factor means before the student calculates the power. A correct first interval checks percentage interpretation directly.
The first interval is a useful model check before a longer power calculation. Ask the student to find the new quantity from the percentage directly and then compare it with the formula at t=1. If they disagree, the factor needs attention. If they agree but the second interval uses the old fixed increase again, teach repeated multiplication. A fresh initial value tests the meaning without changing the percentage. Later, change the percentage itself. Keep the question's stated assumptions visible so an illustrative sequence is not mistaken for a claim that an actual quantity will grow indefinitely.
2. A retained decay fraction
Try first. A quantity starts at 500 and falls by 20 percent per hour under the model. Find it after two hours.
Worked reasoning. The retained factor is 0.8 each hour. Use Q(t)=500(0.8)ᵗ, giving Q(2)=320. The first decrease removes 100, while the second removes 80; equal percentages do not remove equal amounts.
Check. The sequence 500,400,320 has ratio 0.8 in each interval. The quantity stays positive.
Error to notice. Subtracting 100 again gives 300, which follows a fixed-decrease model instead of the stated percentage model.
Independent variant and answer. Starting at 800 and falling 25 percent per interval gives 450 after two intervals.
What this tells the tutor. This check separates model selection from exponent calculation. If the factor is correct, preserve it while inspecting later arithmetic.
A decay factor represents the retained fraction, so it remains positive in these examples. Ask how much of the current quantity is left after one interval. Then compare “falls by” and “falls to” using the same stated percentage. This wording contrast can expose the first modelling fault before logarithms obscure it. A changed initial quantity can test execution, while a changed phrase tests interpretation. The tutor should choose the follow-up that matches the observed need. Do not treat a correct calculator power as evidence that the retained factor was independently understood.
3. A factor belongs to a specified interval
Try first. A model halves every three hours from initial 96. Find Q at six hours.
Worked reasoning. With t in hours, Q(t)=96(1/2)ᵗ⁄³. At t=6 the exponent is 2, so Q=24. The exponent counts three-hour intervals. At t=3 it gives one halving and Q=48.
Check. The six-hour duration contains two stated half-life intervals, matching 96→48→24.
Error to notice. Using exponent six applies six halvings and changes the meaning of the time unit.
Independent variant and answer. Initial 160 halving every four hours gives 40 at eight hours.
What this tells the tutor. Ask which time makes the exponent one. This identifies a unit mismatch before logarithm solving is introduced.
Write the time unit beside t and the interval length beside the exponent denominator. At the stated half-life, the exponent should be one. At twice that duration it should be two. These two checks are simple enough to explain without solving an inverse equation. If the student instead uses the elapsed hours as the number of halvings, preserve their repeated-factor idea while repairing the interval count. A fresh half-life with a different duration tests that boundary. The model's time origin should also be stated so elapsed time is not confused with a clock reading.
4. Recover a factor from exact observations
Try first. A hypothetical model Q=Abᵗ has Q(0)=100 and Q(2)=144. Find A and positive b.
Worked reasoning. The initial condition gives A=100. Then 100b²=144, so b²=1.44 and the positive model factor is b=1.2. The model is Q=100(1.2)ᵗ. Positivity of the chosen exponential factor is part of this model setting.
Check. At t=0 the model gives 100, and at t=2 it gives 144. At t=1 it predicts 120.
Error to notice. Do not infer a per-step factor of 1.44 from a two-step observation. The time span belongs in the exponent.
Independent variant and answer. Q(0)=200 and Q(2)=128 give b=0.8 and Q=200(0.8)ᵗ.
What this tells the tutor. Ask which condition determines A and which determines b. This helps distinguish initial-value reading from elapsed-time factor recovery.
A positive exponential factor is part of the selected model setting. An even-power equation may have two algebraic signs, but a negative base is not appropriate for a real-time exponential model defined through arbitrary real exponents. State why the positive factor is chosen rather than silently discard a sign. Then verify both supplied observations. The tutor can distinguish model restrictions from ordinary quadratic root rules. A follow-up decay observation tests whether the learner recovers a factor below one and recognises that as positive decay, not a negative quantity or rate chosen by guesswork.
CHAPTER 14 OF 17 · See what the work reveals
14. Worked learning checks: meaning and conditions
5. Observations not beginning at zero
Try first. For Q(t)=Abᵗ, Q(1)=60 and Q(3)=135. Find positive A,b.
Worked reasoning. Take the ratio: b²=135/60=2.25, so b=1.5. Then A×1.5=60 gives A=40. The model is Q=40(1.5)ᵗ. The elapsed difference is two, even though the later observation occurs at time three.
Check. At t=1 the model gives 60; at t=3 it gives 40×3.375=135.
Error to notice. Treating the first observation as A would ignore that it occurs at t=1, not zero.
Independent variant and answer. For Q(1)=80 and Q(3)=180, b=1.5 and A=160/3.
What this tells the tutor. This example tests ratio formation and time origin. The tutor should inspect those equations before judging the final numerical parameters.
The ratio uses the elapsed difference between observations. Ask the learner to mark the two observation times and calculate that difference before applying the factor. The first observation need not be the initial value. After recovering b, use one actual observation to find A and check the other. A fresh shifted pair of times tests the time-origin decision more strongly than another pair starting at zero. Parents can bring the two parameter equations; they reveal whether a wrong result came from observation interpretation or from later algebra with a correctly formed ratio.
6. A continuous rate from a ratio
Try first. A model Q=200eᵏᵗ has Q(5)=100. Find k.
Worked reasoning. Substitute the observation: e⁵ᵏ=1/2. Taking natural logarithms gives 5k=ln(1/2), so k=ln(1/2)/5, approximately −0.138629 per stated time unit. The negative rate agrees with decay. Keep the exact logarithmic form for later calculations.
Check. At t=5 the exponential factor is eˡⁿ⁽¹⁄²⁾=1/2, recovering 100.
Error to notice. The rate is not −0.5 simply because the quantity halves. The time and exponential inverse both matter.
Independent variant and answer. For Q=300eᵏᵗ and Q(4)=600, k=ln 2/4.
What this tells the tutor. Ask why the rate sign is sensible and which time unit it uses. These checks support interpretation beyond calculator digits.
The sign of a continuous rate can be predicted from the observation ratio. A decline gives a ratio below one and a negative natural logarithm, hence negative k in Aeᵏᵗ for positive forward elapsed time. If the model uses Ae⁻ᵏᵗ with k positive, the sign convention is different but the quantity can describe the same decay. Read the actual formula before assigning a sign. The tutor should make that convention explicit. A fresh growth ratio tests whether the learner can interpret a positive rate rather than copy a negative sign from the demonstration.
7. Solve an increasing threshold
Try first. For Q(t)=100(1.2)ᵗ, find the real time at which Q=200.
Worked reasoning. Divide by 100 to obtain 1.2ᵗ=2. Therefore t=ln 2/ln 1.2, approximately 3.8018 time units. The factor exceeds one, so the quantity grows and a positive doubling time is expected.
Check. Substituting the exact time makes (1.2)ᵗ=2. At t=3 the quantity is 172.8; at t=4 it is 207.36.
Error to notice. Using 200 rather than the ratio 200/100 inside the logarithm ignores the initial value.
Independent variant and answer. For Q=50(1.5)ᵗ reaching 150, t=ln 3/ln 1.5, about 2.7095.
What this tells the tutor. This tests isolation of the factor and the unknown exponent. Report the unit and distinguish real crossing time from a whole-step request.
Isolate the exponential factor before taking logarithms. The target ratio C/A is the dimensionless multiplier the model must achieve. Ask the student to predict whether it is above or below one and whether the resulting time should be positive within the stated forward domain. Then evaluate the exact logarithm ratio with sufficient precision. A fresh target can fall between two known integer-step values, providing an independent magnitude check. Keep the real crossing time distinct from any later request for a whole-day or whole-step answer.
8. The first whole step satisfying a condition
Try first. For Q(n)=100(1.2)ⁿ at integer n≥0, find the first step with Q≥200.
Worked reasoning. The continuous crossing is about 3.8018, so inspect n=4 and the preceding step. Q(4)=207.36≥200 while Q(3)=172.8<200. The first integer step is 4. Increasing behaviour ensures no earlier nonnegative step can satisfy the threshold.
Check. The neighbouring original-model values verify the discrete answer directly.
Error to notice. Rounding to the nearest integer is not the principle. Choose the earliest allowed step that satisfies the actual comparison.
Independent variant and answer. For Q(n)=100·2ⁿ and Q>400, the first step is 3; step 2 equals 400 and fails strict >.
What this tells the tutor. A strict boundary at an integer tests the reporting decision. The tutor can preserve a correct logarithm time while repairing step selection.
The first permitted step is a membership decision, not ordinary rounding. Check the candidate step and the one before it in the original model. Strictness matters when equality occurs exactly at a permitted step. A monotonic model then justifies that no earlier step has been overlooked. If the student's continuous solution is right but the discrete result fails, preserve the logarithm work and teach this interpretation. A later threshold with exact equality at an integer is a useful focused test because the usual ceiling shortcut needs the comparison symbol to be read carefully.
CHAPTER 15 OF 17 · See what the work reveals
15. Worked learning checks: connecting representations
9. A decay threshold gives a positive time
Try first. For Q(t)=800(0.75)ᵗ, solve Q=200.
Worked reasoning. Isolate 0.75ᵗ=1/4. Then t=ln(1/4)/ln 0.75, approximately 4.8188. Both logarithms are negative, so their ratio is positive. The quantity decreases under the positive factor below one.
Check. At the exact time the factor is 1/4 and the model gives 200. The positive duration agrees with forward decay.
Error to notice. A negative logarithm does not force a negative elapsed time. Inspect the complete ratio and the model domain.
Independent variant and answer. For Q=500(0.8)ᵗ reaching 100, t=ln(0.2)/ln(0.8), about 7.2126.
What this tells the tutor. Ask which sign the elapsed time should have before evaluating. A reasonableness check can reveal an inverted ratio or lost negative sign.
Both logarithms in a decay threshold ratio can be negative, producing a positive time. Ask the learner to inspect the complete quotient rather than infer its sign from the numerator alone. An inverted target ratio or a missing denominator sign can produce an implausible negative elapsed value. The original-model substitution is the final check. A fresh retained fraction and target ratio test the same boundary without repeating the previous numbers. If the mathematical answer lies outside a stated time domain, explain that interpretation rather than force the result to fit a preferred positive value.
10. An excess above a baseline
Try first. A hypothetical model T(t)=20+60e⁻ᵏᵗ with k>0 has T(5)=50. Find k and T(10).
Worked reasoning. At five units, 50−20=60e⁻⁵ᵏ, so e⁻⁵ᵏ=1/2 and k=ln 2/5. At ten units the excess has halved twice, giving T(10)=20+60/4=35. The total starting value is 80, while the initial excess is 60.
Check. The model gives T(0)=80, T(5)=50 and T(10)=35. Its excess sequence is 60,30,15.
Error to notice. Halving the total 80 would give 40 at five units, inconsistent with the stated baseline model.
Independent variant and answer. For T=10+80e⁻ᵏᵗ with T(4)=50, k=ln 2/4 and T(8)=30.
What this tells the tutor. This checks baseline subtraction before logarithms. The student should identify which quantity follows the multiplicative decay.
A baseline model has a constant part and a changing excess. Label the initial total and initial excess separately. The first observation equation should subtract the baseline before forming a ratio. Ask the learner to identify which quantity halves over the stated duration. A second duration then gives another halving of that excess, followed by restoring the baseline. This tests the full route. The tutor should not infer that a learner who knows the half-life formula can automatically interpret an offset model; the offset adds a separate meaningful decision.
11. A target the model never reaches
Try first. For T(t)=20+60e⁻ᵏᵗ with k>0 and t≥0, can T equal 20 at finite time?
Worked reasoning. No. The exponential factor is strictly positive at every finite real t, so T>20. As t increases without bound, the exponential contribution approaches zero and T approaches 20. A target below 20 is also impossible in this positive-excess model.
Check. Setting T=20 would require e⁻ᵏᵗ=0, which has no finite real solution. The conclusion follows from the model structure.
Error to notice. Do not treat a very small displayed exponential value as exactly zero or a finite arrival at the baseline.
Independent variant and answer. For Q=5+30e⁻ᵗ, the model remains above 5 at finite time and approaches 5.
What this tells the tutor. This tests interpretation and domain rather than numerical solving. Ask whether the requested target is reachable before applying a logarithm to a zero ratio.
Reachability can be decided from the model's structure before inverse calculation. A strictly positive exponential contribution never becomes exactly zero at finite real time, even when a calculator display rounds it to zero. Ask what target ratio would be required for the baseline and whether it is allowed by the logarithm. This connects domain and model interpretation. A fresh target below the baseline should also be rejected for a reason. The learner should not press ahead with a logarithm of zero or a negative excess simply because previous examples always produced a numerical time.
12. Compare a difference model with a ratio model
Try first. Two hypothetical models start at 100: L(t)=100+20t and E(t)=100(1.2)ᵗ. Compare at t=1 and t=2.
Worked reasoning. At one unit both give 120. At two units the linear model gives 140 and the exponential gives 144. One common observation does not distinguish their later behaviour. The linear model has fixed difference 20 per unit; the exponential has fixed factor 1.2.
Check. At t=3 the values are 160 and 172.8, confirming their distinct rules. The comparison is mathematical under the stated models.
Error to notice. Do not infer a unique exponential rule from one matching step without additional information or an explicit model assumption.
Independent variant and answer. Starting at 200, compare 200+40t with 200(1.2)ᵗ: at t=2 they give 280 and 288.
What this tells the tutor. This final contrast tests the core modelling decision before more complicated parameter or threshold calculations are attempted.
A shared first observation does not uniquely identify a model. Compare the second step through each stated rule: one adds a fixed amount, the other multiplies by a fixed factor. Then ask what information in the actual question supports choosing a model. In a classroom problem it may be explicitly specified; in a genuine data situation more evidence and assumptions are needed. Keep that distinction honest. A later comparison with changed starting quantities tests whether the learner recognises the rule rather than remembers one familiar sequence. Model choice should precede parameter solving, not be inferred from a convenient final decimal.
Does a 15 percent increase use factor 0.15?
No. The retained whole plus the increase gives 1.15 per stated interval.
Is decay represented by a negative quantity?
A positive decay model usually uses a factor between zero and one or a negative exponential rate. Its output can remain positive.
Can we use the same rate with different time units?
The numerical rate must be converted so the exponent represents the same elapsed time.
Does half-life mean the total always halves?
In a simple zero-baseline exponential it does. With a nonzero baseline, identify whether the excess above that baseline is the quantity that halves.
Should a first whole-day threshold be rounded normally?
Find the first permitted step that actually satisfies the original condition, checking it and the preceding step.
What should we show a tutor?
Bring the model line, time definition, initial value, ratio or percentage interpretation and original-model checks.
Secondary 3 Additional Mathematics Tuition: Why Are Quadratic Inequalities Different from Equations?
Secondary 3 Additional Mathematics Tutor: Why Do Simultaneous Equations Sometimes Give Two Answers?
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