To find when two exponential models become equal, set their expressions equal, divide by positive terms, and use logarithms to solve the exponent. If A(t) = 200(6/5)^t and B(t) = 320(21/20)^t, then equality means (8/7)^t = 8/5, so the exact crossing time is t = ln(8/5)/ln(8/7), approximately 3.52 time units.
This is a different question from finding the time when one model reaches a fixed threshold. Both quantities are changing, perhaps at different rates. The key is to compare their relative multipliers rather than treating one side as a constant. This Additional Mathematics guide explains crossover times, first whole-number periods, growth versus decay, models with no intersection and checks for meaningful domains.
What a crossover time means
Two functions cross when they have the same output at the same input. If time t is measured in days, the equality A(t) = B(t) asks for a day or fractional day when the two modelled quantities agree.
A comparison may represent two hypothetical savings plans, two populations, two inventory projections or two mathematical growth curves. In each case, the models must use compatible units and the same time origin to make the comparison meaningful.
The crossover time is not necessarily an integer. A continuous model can predict equality between recorded daily measurements, while a problem about the first whole day of overtaking requires a separate rounding or adjacent-integer check.
The first decision is to distinguish exact equality from an inequality such as A(t) > B(t). Solving the equality identifies the boundary; the inequality tells us which model is larger on either side, provided both models remain within their permitted domain.
Read the two multipliers accurately
For A(t) = 200(6/5)^t, the starting amount is 200 and the per-unit growth factor is 6/5 = 1.2. For B(t) = 320(21/20)^t, the starting amount is 320 and the per-unit growth factor is 21/20 = 1.05.
Thus B starts larger at t = 0, but A grows faster in percentage terms: 20% versus 5% each time unit. A crossover is plausible because the initially smaller model has the greater growth factor.
The starting difference is 120, but the future behaviour cannot be understood by repeatedly adding fixed amounts. Exponential changes multiply the current value, making each successive absolute increase depend on what the quantity has already become.
A qualitative prediction should come before algebra: if both models continue indefinitely with positive factors and A grows faster, the ratio A(t)/B(t) increases. That predicts at most one crossover when their multipliers differ.
Set the models equal and simplify first
Begin with 200(6/5)^t = 320(21/20)^t. Divide both sides by the positive number 200(21/20)^t.
The result is [(6/5)/(21/20)]^t = 320/200. Simplify the base ratio: (6/5)(20/21) = 120/105 = 8/7. Simplify the output ratio: 320/200 = 8/5.
Thus the equality is equivalent to (8/7)^t = 8/5. The two original exponential expressions have become a single exponential equation with a positive right-hand side.
This division is valid because positive exponential bases produce positive outputs for real exponents. A good solution performs the ratio simplification before taking logarithms, which often produces an exact expression more readable than a line of unrelated decimal approximations.
Solve the exponent with logarithms
For the equation (8/7)^t = 8/5, take natural logarithms of both sides: t ln(8/7) = ln(8/5).
Because 8/7 is not 1, its logarithm is non-zero. Therefore t = ln(8/5)/ln(8/7), approximately 3.52.
Any consistent real logarithm base greater than zero and different from one gives the same ratio. Common logarithms would yield log(8/5)/log(8/7). The natural logarithm is simply a convenient choice.
Keep the exact logarithmic form through the working. A rounded t-value does not make the models exactly equal when substituted; it is an approximation to the continuous crossover time. State the approximation only to the precision requested.
Check which model is larger before crossing
At t = 0, A(0) = 200 and B(0) = 320, so A is smaller.
At t = 3, A(3) = 200(6/5)^3 = 345.6, while B(3) = 320(21/20)^3 = 370.44. B remains larger.
At t = 4, A(4) = 200(6/5)^4 = 414.72. Meanwhile B(4) = 320(21/20)^4 = 388.962. A has overtaken B.
The comparison agrees with a crossover between t = 3 and t = 4. The exact logarithmic solution gives about 3.52. These integer checks provide a meaningful interpretation instead of leaving a logarithm ratio detached from the modelled situation.
First whole-number period means a different final answer
If a question asks for the first whole-number t when A(t) exceeds B(t), the exact continuous crossing at approximately 3.52 is an intermediate boundary, not the final integer answer.
The ratio A(t)/B(t) increases with t because the relative multiplier 8/7 exceeds one. At t = 3 the ratio is still below one; at t = 4 it is above one.
Therefore the first non-negative integer t with A(t) > B(t) is 4. Do not round 3.52 to the nearest integer by habit. Instead identify the smallest permitted integer strictly greater than the boundary and verify it in the inequality.
If equality happened at an integer, then a strict overtaking condition would require the next integer, while a non-strict “at least as large” condition could include the equality time itself. Always read the comparison sign.
Why the relative ratio is the key mathematical object
Define R(t) = A(t)/B(t) when both quantities are positive. For the worked models, R(t) = (200/320)[(6/5)/(21/20)]^t = (5/8)(8/7)^t.
Equality occurs when R(t) = 1. The model A is larger when R(t) > 1 and smaller when R(t) < 1.
Because 8/7 > 1, the ratio grows strictly with time, even though both individual models also grow. This establishes that there can be only one real crossover.
This approach is especially useful when the starting values and rates are awkward. Comparing one exponential ratio is generally easier than imagining two curves separately and trying to guess where they meet on a rough sketch.
A general exact crossover formula
Let A(t) = P a^t and B(t) = Q b^t with P, Q, a and b all positive. Equality requires P a^t = Q b^t.
Divide to obtain (a/b)^t = Q/P. If a ≠ b, take logarithms and solve: t = ln(Q/P)/ln(a/b).
The formula does not guarantee that the time lies in the practical model’s allowed interval. For a problem restricted to t ≥ 0, a negative calculated time describes a crossing before the chosen start and is not an allowed future crossover.
The derivation is often more useful than memorising the formula. Recreating the ratio step makes the roles of starting amounts, per-unit multipliers and time constraints clearer and helps avoid reversing one of the ratios accidentally.
What if the crossing time is negative?
Suppose A(t) = 400(6/5)^t and B(t) = 200(21/20)^t. A begins twice as large and also grows faster.
Equality would require (8/7)^t = 1/2, giving t = ln(1/2)/ln(8/7), a negative number. The two mathematical curves intersect only at a negative time relative to this chosen origin.
If the model applies only for t ≥ 0, there is no allowed crossover in the modelled period. A(t) remains larger throughout that domain.
This is not an algebraic failure. It is a domain interpretation. A valid real solution of an unrestricted equation need not be a valid answer to a question about non-negative elapsed time.
When two models have identical multipliers
Suppose A(t) = 200(11/10)^t and B(t) = 320(11/10)^t. Their multiplier is identical, so their ratio is always 200/320 = 5/8.
They never become equal because the starting amounts differ and both are scaled by the same positive factor at every time. Dividing the equality would give 200 = 320, a contradiction.
The general logarithmic formula is not applicable here because ln(a/b) = ln 1 = 0, and division by zero is undefined.
If the starting amounts also agree, the two models are identical for every t, so every time in their common domain is an equality time. Recognise these cases before applying a formula that assumes unequal growth factors.
Growth versus decay can still have one crossover
Suppose A(t) = 800(4/5)^t is decaying and B(t) = 100(11/10)^t is growing, for t ≥ 0.
Their equality becomes 800(4/5)^t = 100(11/10)^t. Divide to obtain 8 = [(11/10)/(4/5)]^t = (11/8)^t.
Thus t = ln 8/ln(11/8), approximately 6.53. The declining model begins larger, while the growing one eventually overtakes it.
The same algebra works for growth–growth, decay–decay and growth–decay comparisons, provided both bases and starting amounts are positive. What changes is the sign of the logarithms and the direction of the models’ relative ratio.
Both models can decay at different speeds
Consider A(t) = 500(9/10)^t and B(t) = 400(19/20)^t. A starts higher but decays by 10% each interval; B decays by only 5%.
Equality gives [(9/10)/(19/20)]^t = 400/500, or (18/19)^t = 4/5. Both sides lie between zero and one.
The exact crossover is t = ln(4/5)/ln(18/19), a positive number because both logarithms are negative.
A slower-decaying model can overtake a faster-decaying model even when neither is growing. The correct comparison is the relative multiplier 18/19, not just the fact that both functions decrease over time.
The role of time units
The multiplier a in P a^t is tied to the unit used for t. If t counts years, a is the yearly multiplier; if t counts months, it must be converted appropriately.
For instance, a 20% yearly growth model written P(6/5)^t becomes P[(6/5)^(1/12)]^m when m = 12t counts months.
A comparison of two models expressed in different units is invalid until both exponents refer to the same time measure. One cannot directly compare a yearly multiplier of 1.2 with a monthly multiplier of 1.05 as though they represented equal-duration intervals.
Write the units beside each t-variable and convert both models before taking their ratio. A numerically correct logarithmic solution in inconsistent units has no meaningful practical interpretation.
The importance of a common time origin
Two exponential models may use different reference dates. For example, one model may start at t = 0 while another is written using t − 2.
To compare at the same actual time, first rewrite both expressions in one common variable. The equality condition must evaluate both quantities at the same event or elapsed time.
Suppose A(t) = 200(6/5)^t and B(t) = 320(21/20)^(t−2). The second model’s stated 320 is its value at t = 2, not at t = 0.
It would be incorrect to use the starting ratio 320/200 without accounting for the shift. Rearrange the shifted model as B(t) = [320/(21/20)²](21/20)^t before applying the ordinary crossover formula.
Why a fixed baseline complicates the algebra
Some practical models have an added baseline, such as A(t) = C + P a^t. If two models share the same C, the baseline cancels when the models are set equal, and a simple exponential ratio may remain.
If their baselines differ, the equality C₁ + P a^t = C₂ + Q b^t generally cannot be reduced to one exponential by merely dividing. The difference of constants remains.
Such a problem may require another algebraic transformation, a graph or a numerical method, depending on the supplied structure and syllabus.
Do not take logarithms across a sum or difference as though log(A + B) equalled log A + log B. That is not a valid logarithm law and can produce a very plausible-looking incorrect crossing time.
A graph helps confirm the direction of overtaking
For the main example, plot the two outputs at t = 0, 3 and 4. B starts at 320 while A starts at 200. By t = 3 their values are close, and by t = 4 their order has reversed.
A rough sketch should show the faster-growing A curve crossing the slower-growing B curve exactly once at t ≈ 3.52.
This is a reasonableness check rather than a method of obtaining the exact logarithmic answer. A plot may not be sufficiently precise to distinguish 3.50 from 3.52.
Where a graphing tool is available, use it to inspect the solution, not to replace the equation and its domain. The algebra demonstrates uniqueness and allows a clear distinction between continuous equality and discrete overtaking.
Why a fixed absolute difference is misleading
At t = 0 the difference B − A is 120. One might be tempted to compare the original percentage growth rates and subtract a fixed amount each day until the gap disappears.
But the absolute increases depend on the current model values. During the first interval, A grows by 40 while B grows by 16. During later intervals both increases change.
The difference B(t) − A(t) is therefore not generally linear. It should not be extrapolated using one day’s reduction as though the same difference closes at a constant rate.
Dividing the two models replaces this awkward difference with a single relative multiplier, which captures how the comparison evolves over equal intervals. That is the right structure for finding an exact crossover.
A check using logarithms before evaluating decimals
From A(t) = B(t), the exact equation is t ln(8/7) = ln(8/5). This already shows t > 0 because both logarithms are positive.
It also shows t is between 3 and 4 by comparing powers: (8/7)^3 < 8/5 < (8/7)^4. Thus an approximate answer in that interval is reasonable before any calculator evaluation.
Only after establishing the exact form should a student calculate the decimal time. Retain sufficient precision if the result is later used in another formula.
This helps prevent a common numerical issue: rounding the intermediate ratio or logarithm too early and then reporting a crossover value inconsistent with the original exponential equality.
Do not report negative model values
A model P a^t with P > 0 and a > 0 is positive for every real t. Therefore two such models can meet only at a positive output, even when the crossing time is negative.
If a question claims a negative crossover amount under these assumptions, the statement is incompatible with the model or the equation has been set up incorrectly.
A separate model with an additive baseline may take negative values, but that is a different formula and requires different interpretation.
This positivity check is a quick conceptual safeguard. Exponential and logarithmic calculations must respect their input and output domains, not merely be manipulated as strings of symbols.
A focused examination solution format
Write both models with the same time variable and units. State the allowed domain, such as t ≥ 0, before solving.
Set the expressions equal, divide by positive terms and simplify the ratio of multipliers. Take logarithms to obtain the exact crossing time.
Check whether the result lies in the allowed time domain. If the question asks which model is larger or for the first whole-number interval, evaluate the relative ratio or adjacent whole-number times.
Report the requested output precisely: an exact continuous t, a suitably rounded time or a first admissible integer. These are different mathematical answers to closely related questions.
Practice questions with checked answers
Question one: A(t) = 200(6/5)^t and B(t) = 320(21/20)^t. Find the continuous equality time. Answer: ln(8/5)/ln(8/7), about 3.52.
Question two: with the same models and non-negative integer t, find the first whole t with A(t) > B(t). Answer: t = 4.
Question three: A(t) = 800(4/5)^t and B(t) = 100(11/10)^t. Find equality time. Answer: ln 8/ln(11/8), about 6.53.
Question four: A(t) = 200(11/10)^t and B(t) = 320(11/10)^t. When are they equal? Never, because their ratio is fixed at 5/8.
Question five: A(t) = 400(6/5)^t and B(t) = 200(21/20)^t for t ≥ 0. Is there a future crossover? No; the unrestricted equality time is negative.
For each, verify the initial ordering and the multiplier ratio before using logarithms.
Questions families often ask
Does the faster percentage growth model always start larger? No. It may begin smaller and overtake later, which is what the example studies.
Is the crossing time necessarily a whole day? No. The continuous time can be fractional. A separate first-whole-day question needs an integer check.
Why use logarithms? The unknown time appears in an exponent, and logarithms reverse exponential operations.
Can two different models never cross? Yes. Equal multipliers with different starting amounts never meet; a crossing may also lie outside the permitted time domain.
How do we check the answer? Evaluate both models near the estimated crossing and compare their ratio, keeping the exact logarithmic result until the final calculation.
Learning transfer, syllabus and next reading
Crossover problems link exponential functions, logarithm laws, inequalities and mathematical modelling. Some comparison scenarios are enrichment depending on the learner’s particular school and assessed syllabus.
For Singapore, consult the appropriate SEAB lists: 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232. Match practice difficulty to the school sequence.
For further reading, see How to Reconstruct an Exponential Model from Two Observations, How to Find the First Whole Day an Exponential Threshold Is Crossed and Why a Logarithmic Inequality Reverses When Its Base Is Below One. Explore the Additional Mathematics archive or enquire about eduKate Singapore’s small-group Mathematics tutorials near Sixth Avenue MRT.
The objective is to compare two changing processes using valid algebra and report the result in the time domain the question actually asks about. Properly taught kids shine a bright light into the future.