Exponential and log models

Initial value, rate constant, value, time

In 9 of 20 recent exams
2% of the marks
Paper 2, calculator
1 to 3 marks

How it is set in the exam

Paper 2 (calculator) in every 2021–26 instance: 1–3 marks a part, or one unstructured part of up to 7 marks; a whole Section A question (5–7 marks), or the first parts of a Section B question that then moves to rates or a second model. Common (9 of 20 exam versions, 31 parts).

Spot it in the exam

Exam phrases to spot (7)

“the mass remaining after $t$ days is modelled by $M(t) = M_0e^{-kt}$ ”
“show that $k = \frac{\ln 2}{T}$ ”
“the half-life of the substance is …”
“find the time taken for the amount to fall to …”
“after how many complete minutes is … first below …”
“interpret the value $1.035$ in this context”
“give a reason why the second model is more realistic”

In disguise (6)

calendar years as $t$ : " $t$ years after the start of 2016", then a question about 2027
a percentage decayed or lost: "40% has decayed" means $60\%$ remains
a model that levels off, $c + b\,e^{-kt}$ (cooling, a background level): the start is $c + b$ , and $c$ is the level it settles at
a second, bounded model given in full: it is not a family to learn, only a formula to substitute into and solve
a log-scale reading, such as a pH or a loudness in decibels: evaluate it, or go backwards to the input
"write the model in the form $N_0e^{kt}$ ": $k = \ln b$ , exactly

Variants (3)

the model: decay $N_0e^{-kt}$ , growth $N_0e^{kt}$ , levelling off $c + b\,e^{-kt}$ , a growth factor $N_0b^t$ , a bounded model given in full, a log scale
the fact that fixes a constant: a half-life, a halving time, a percentage change between two dates, a percentage decayed, one data point
what is asked: the initial value, $k$ (often a show-that), a value, a time, the first whole unit, the meaning of a constant, a reason one model is better

Looks like it, but isn't (8)

a fixed percentage per step and no formula in continuous time: a geometric sequence, Find n or Compound interest and depreciation
how fast the model is changing, or its long-run value or long-run rate found as a limit: Rate of change in context. Saying what a given constant represents (the level $c$ ) stays here
when a model is growing fastest (the greatest rate): Concavity and points of inflexion
the total amount from a given rate: Total change from a rate
an exponential equation with no context and no data: Solve an exponential equation
when two models are equal: Solve graphically; the greatest difference between two models: Stationary points
a periodic model (tides, a wheel): a sinusoid, in Trig graphs and sinusoidal models
a straight line fitted to paired data, with $r$ : Regression on the GDC. Fitting an exponential curve to data (exponential regression) is AI; in AA an exponential model is given, or its constants are found from facts stated in the question

Worked example

A half-life: show that k, then a percentage broken down

Paper 2 (calculator)

Question

A farmer sprays a field with a pesticide. The amount of pesticide in each kilogram of soil $t$ days later is modelled by $M(t) = M_0e^{-kt}$ milligrams, where $k > 0$ . When $t = 0$ there are $45$ mg in each kilogram, and the half-life of the pesticide is $12$ days.
(a) Write down the value of $M_0$ . [1]
(b) Show that $k = \dfrac{\ln 2}{12}$ . [3]
(c) Find the amount of pesticide in each kilogram of soil after $30$ days. [2]
(d) Find the number of days, to the nearest day, until $90\%$ of the pesticide has broken down. [3]

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See it move: Two points fix the model

Through $(0, 45)$ and $(12, 22.5)$ : $22.5 = 45e^{-12k}$ , so $e^{-12k} = 0.500$ and $k = -\frac{\ln 0.500}{12} = 0.0578$ per day. The half-life is $\frac{\ln 2}{k} = 12.0$ days.
(a) At $t = 0$ , $e^{0} = 1$ : $M(0) = M_0 = 45$ , the start, read off exactly.
(b) The half-life is one fact for the one unknown: after 12 days half of 45 is left. $22.5 = 45e^{-12k}$ , $e^{-12k} = \frac{1}{2}$ , $-12k = \ln\frac{1}{2} = -\ln 2$ , so $k = \frac{\ln 2}{12}$ . A "show that" wants these steps, not a GDC value.
(c) $M(30) = 45e^{-0.0578 \times 30} = 7.95$ mg per kg: 30 days is 2.5 half-lives, so it sits between $\frac{45}{8} = 5.625$ and $\frac{45}{4} = 11.25$ .
The level $M = 4.5$ : $45e^{-kt} = 4.5$ , $e^{-kt} = 0.100$ , so $t = \frac{\ln(45/4.5)}{k} = 39.86$ days: 40 to the nearest day. The whole days either side: $M(39) = 4.73$ and $M(40) = 4.46$ ; a "least number of whole days" rounds up to 40.
(d) "90% broken down" means 10% remains: the equation is $M(t) = 4.5$ , not $M(t) = 40.5$ . After 3 half-lives (36 days) 12.5% is left and after 4 (48 days) 6.25%, so the answer lies between 36 and 48.

Which method?

a constant such as $k$ , or a time, on
Method
isolate the exponential and take $\ln$ ; or the GDC (solve an equation, zeros and intersections)
Write down
the equation with the fact substituted, then its solution
"show that $k = \ldots$ "
Method
$\ln$ by hand, forwards, with at least two log steps seen; the GDC can't show it
Write down
every step down to the given form
"after how many complete …", "the first whole …"
Method
solve the equation, then round up; or a table of whole-number values (table of values)
Write down
the solution, then the values at the whole numbers either side
"in which year …"
Method
solve for $t$ ; when $t = 0$ is the start of a year, the answer is that year plus the whole part of $t$ : with $t = 0$ at the start of 2016, $t = 21.3$ is during $2016 + 21 = 2037$ (not $2016 + 22$ )
Write down
$t$ , then the year
a constant of a model given in full
Method
the GDC: solve an equation
Write down
the equation

The part asks for	Method	Write down
a constant such as $k$ , or a time, on	isolate the exponential and take $\ln$ ; or the GDC (solve an equation, zeros and intersections)	the equation with the fact substituted, then its solution
"show that $k = \ldots$ "	$\ln$ by hand, forwards, with at least two log steps seen; the GDC can't show it	every step down to the given form
"after how many complete …", "the first whole …"	solve the equation, then round up; or a table of whole-number values (table of values)	the solution, then the values at the whole numbers either side
"in which year …"	solve for $t$ ; when $t = 0$ is the start of a year, the answer is that year plus the whole part of $t$ : with $t = 0$ at the start of 2016, $t = 21.3$ is during $2016 + 21 = 2037$ (not $2016 + 22$ )	$t$ , then the year
a constant of a model given in full	the GDC: solve an equation	the equation

Steps

Read the model:
what $t$ measures, where $t = 0$ is, the units, and the form ( $N_0e^{kt}$ , $N_0b^t$ , $c + b\,e^{-kt}$ , given in full, or a log scale).
Initial value:
$t = 0$ , and $e^0 = 1$ . It is exact: $c + b$ for a levelling-off model.
A constant from one fact:
substitute the fact (half the initial value for a half-life; $0.82 \times$ the start for "fell by 18%"; $60\%$ for "40% decayed"; a data point). Isolate the exponential, then take $\ln$ , or use the GDC. Store the unrounded value (keep full-accuracy values).
A value at a time:
turn dates into $t$ first. Write the substitution, e.g. $N(11) = 2400 \times 1.035^{11}$ , then the value (define a function and evaluate it). A count of people, animals or trees is a whole number.
The time for a level:
set the model equal to the level (an inequality, such as $C(t) < 800$ , scores the same) and solve. For the first whole unit after the level is passed ("after how many complete minutes …", "the least number of whole …"), round up and check the values at the whole numbers either side. (For how many complete units a model stays above or below a level, round down.)
Interpret:
the growth factor $b$ of $N_0b^t$ , $b = 1 + r$ , is "grows by $r \times 100\%$ per year" (the unit of time is part of the answer); $c$ in $c + b\,e^{-kt}$ is the level it settles at; a model is more realistic if it respects a limit in the context (a fixed number of people, a room temperature).
A log-scale model:
substitute to evaluate. To go back to the input, divide by the coefficient before anything else, then write the log as a power: $2\log_{10} X = 7$ gives $\log_{10} X = 3.5$ , so $X = 10^{3.5}$ . An exact power of $10$ is accepted. An input multiplied by $m$ changes the reading by the coefficient times $\log_{10} m$ (the Log laws).

Common mistakes

A percentage decayed used as the amount left
" $90\%$ has broken down" leaves $10\%$ . Markschemes give a mark for the amount remaining.
Halving the wrong thing
in $c + b\,e^{-kt}$ : "half its initial value" is half of $c + b$ , not half of $b$ .
Reading $k$ as the percentage rate
$e^{-0.05t}$ is a fall of $4.88\%$ per unit of time, not $5\%$ . The rate comes from the factor per unit of time, $e^k$ .

More mistakes (12)

Rounding a time to the nearest whole unit
when the first whole unit is asked: $12.2$ minutes means $13$ , not $12$ (and $12.9$ also means $13$ ).
Ignoring the rounding the question asks for
"to the nearest day" needs $40$ , not $39.9$ ; the final mark is for the form asked for.
Taking $\ln$ before isolating the exponential
$\ln\left(400 + 1400e^{-10k}\right)$ does not split. Move the constant across and divide by the coefficient first: $e^{-10k} = \frac{5}{14}$ .
Undoing a log scale in the wrong order
$2\log_{10} X = 7$ gives $\log_{10} X = 3.5$ first, so $X = 10^{3.5}$ ; writing $X = \frac{10^{7}}{2}$ is wrong. Divide by the coefficient before writing the log as a power.
Putting the calendar year into the formula
the start of 2027 is $t = 11$ when $t$ counts from the start of 2016.
A value of $t$ where a year is asked
$t = 21.3$ is not an answer to "in which year". Count from $t = 0$ : $2037$ .
A rounded constant carried forward
$q = 0.273$ gives $3273$ people instead of $3272$ , and near a whole-number cut-off a shift like that changes the answer. Store the full value, and write $4$ or $5$ significant figures in the working.
An interpretation with the unit of time missing
" $3.5\%$ " is not enough; " $3.5\%$ per year" is.
A count given as a decimal
$3503.9$ residents is not an answer.
A GDC value in a "show that"
$k = 0.0578$ from the calculator does not show $k = \frac{\ln 2}{12}$ .
A "show that" done backwards
substituting $k = \frac{\ln 2}{12}$ and checking that $M(12) = 22.5$ proves nothing. Start from $22.5 = 45e^{-12k}$ .
A decimal where an exact $k$ is asked
$k = 0.0344$ is not $\ln 1.035$ . Substituting a value from the calculator, such as $2400e^{11k} = 3503.9\ldots$ , gives only a decimal, and markschemes give it no marks. Match the factors for every $t$ : $e^k = 1.035$ .

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