Questions on log graphs and exponential models

Question 1

A quantity is believed to follow a power law y = axⁿ. State what should be plotted to test the claim, and what the gradient and intercept of the resulting line give.

Accepted Answer

Plot log y against log x. Taking logs of y = axⁿ gives log y = log a + n log x, so the points fall on a straight line with gradient n and intercept log a. Straightness on logged axes is the evidence for the law. B1 for what to plot, B1 for the gradient, B1 for the intercept. The intercept is log a, not a, so a is recovered by undoing the logarithm.

Question 2

A quantity is believed to follow exponential growth y = ab^x. State what should be plotted, and what the gradient and intercept give.

Accepted Answer

Plot log y against x itself rather than log x. The law gives log y = log a + x log b, a line with gradient log b and intercept log a. B1 for what to plot, B1 for the gradient, B1 for the intercept.

Question 3

A data set is plotted two ways: log y against log x gives a curve, and log y against x gives a straight line. State which of the two models, y = axⁿ or y = kb^x, fits the data, and explain your reasoning.

Accepted Answer

The model is y = kb^x, the exponential law. B1 for identifying it. A power law only becomes a straight line when log x is used, since log y = log a + n log x is linear in log x: a curve on that plot rules a power law out. B1 for that reasoning. An exponential law becomes a straight line when the raw x is used instead, since log y = log k + x log b is linear in x itself, which matches the second plot exactly. B1 for that reasoning.

Question 4

Explain why testing a power law y = axⁿ against data requires x > 0 as well as y > 0, and state what happens to a measurement taken at x = 0 on a log-log plot.

Accepted Answer

log x is undefined for x ≤ 0, since no power of a positive base produces zero or a negative number, so log x can only be formed when x is positive; the same is true of log y and y. B1 for the positivity condition on both variables. A measurement at x = 0 has no logarithm to plot at all, so that point cannot appear on a log-log axis and has to be left out of the fit, whatever it showed on the original axes. B1 for the point being unplottable, B1 for it being dropped rather than plotted at some extreme position.

Question 5

A quantity follows y = axⁿ, and measurements give (4, 40) and (9, 135). Find n and a, and state the law.

Accepted Answer

The log-log gradient is n = (log 135 − log 40)/(log 9 − log 4) = log 3.375 ÷ log 2.25 = 1.5 exactly, since 3.375 = 2.25^1.5. Then a = 40/4^1.5 = 40/8 = 5, and the law is y = 5x^1.5. M1 for forming the log-log gradient, A1 for n = 1.5, M1 for substituting a point to find a, A1 for the law. The second point checks it: 5 × 9^1.5 = 5 × 27 = 135.

Question 6

A mass in grams is modelled by m = 60e^−0.1t, with t in days. State the initial mass, find the mass after 5 days, and find the half-life of the decay, each to 3 significant figures where rounding is needed.

Accepted Answer

The initial mass is the front constant, 60 g, since e⁰ = 1. After 5 days, m = 60e^−0.5 = 36.4 g. Halving means e^−0.1t = ½, so −0.1t = ln ½ and t = ln 2/0.1 = 6.93 days. B1 for 60 g, M1 for substituting t = 5, A1 for 36.4 g, M1 for setting the exponential to ½ and taking logs, A1 for 6.93 days. The half-life never depends on the starting mass, so the 60 cancels before the log is taken.

Question 7

The model y = 3 × 5^x is to be drawn as a straight line. State what should be plotted, and give the line's gradient and intercept to 3 significant figures.

Accepted Answer

Plot log y against x. Taking logs: log y = log 3 + x log 5, so the gradient is log 5 = 0.699 and the intercept is log 3 = 0.477. B1 for what to plot, B1 for 0.699, B1 for 0.477.

Question 8

A graph of log₁₀ y against x is a straight line through (0, 2) and (10, 3.5). Find the model y = kb^x, giving k and b each to 3 significant figures.

Accepted Answer

The intercept, at x = 0, is log k = 2, so k = 10² = 100. M1 for reading the intercept, A1 for k = 100. The gradient is (3.5 − 2)/(10 − 0) = 0.15, and the gradient is log b, so b = 10^0.15 = 1.41 to 3 significant figures. M1 for the gradient, A1 for b = 1.41.

Question 9

An investment is modelled by V = 500 × 1.08^t, t in years. Using the linear relationship log V = log 500 + t log 1.08, find, to 3 significant figures, the number of years for the investment to reach £2000.

Accepted Answer

Substituting V = 2000 into the linear relationship, log 2000 = log 500 + t log 1.08, so t = (log 2000 − log 500)/log 1.08 = log 4/log 1.08, since 2000/500 = 4. M1 for the linear relationship with V = 2000, A1 for the equation solved for t. This gives t = 18.0 years to 3 significant figures. M1 for the division, A1 for 18.0.

Question 10

A population is modelled by P = 400 × 1.15^t, t in years. Evaluate P at t = 50 to 3 significant figures, and explain, giving a reason grounded in the real system, why the model is unlikely to remain valid that far into the future.

Accepted Answer

P(50) = 400 × 1.15⁵⁰ = 433 000 to 3 significant figures. M1 for substituting t = 50, A1 for 433 000. The model assumes the population keeps growing by the same 15% every year forever, but a real population is limited by food, space and other resources, so the growth rate falls as the population approaches whatever the environment can support, and an unlimited exponential model stops describing it well before it reaches numbers this large. B1 for naming the limiting resource, B1 for stating that the growth rate itself would have to fall.

Question 11

The table gives measurements of a quantity y against x: (1, 3), (4, 6) and (9, 9). By finding log x and log y for each pair, or otherwise, determine whether a power law y = axⁿ or an exponential law y = kb^x fits the data, and state the law.

Accepted Answer

Testing the power law first: the log-log gradient between (1, 3) and (4, 6) is (log 6 − log 3)/(log 4 − log 1) = log 2/log 4 = 0.5, and between (4, 6) and (9, 9) it is (log 9 − log 6)/(log 9 − log 4) = 0.5 again, the same value both times. M1 for forming a log-log gradient, A1 for the two gradients agreeing at 0.5. Testing the exponential law instead, the log-linear gradient between the same pairs is (log 6 − log 3)/(4 − 1) ≈ 0.100 and (log 9 − log 6)/(9 − 4) ≈ 0.0352, which do not agree. M1 for showing the log-linear gradient is inconsistent. So the data follows the power law, not the exponential one. A1 for that conclusion. With n = 0.5, a = 3/1^0.5 = 3, so the law is y = 3√x, and the third point confirms it: 3 × √9 = 3 × 3 = 9. A1 for the law. A single log-log gradient could arise from either family by coincidence at one pair of points; agreement across two independent pairs is what rules the exponential law out here.

Question 12

A log-log plot of some data gives a straight line with gradient 0.5 and intercept log 6. Find the law connecting y and x, and the value of y when x = 25.

Accepted Answer

The line says log y = 0.5 log x + log 6, so the law is y = 6x^0.5 = 6√x. At x = 25, y = 6 × 5 = 30. M1 for undoing the logs, A1 for the law, M1 for substituting x = 25, A1 for 30.

Question 13

For the decay model m = 60e^−0.1t, show that the ratio m(t + 7)/m(t) is the same for every t, and find its value to 3 significant figures.

Accepted Answer

m(t + 7)/m(t) = e^−0.1(t+7)/e^−0.1t = e^−0.7 = 0.497, with the t-dependence cancelling entirely. M1 for forming the ratio, A1 for the t-dependence cancelling, A1 for 0.497.

Question 14

By taking logarithms of y = kb^x, show that the gradient of the line obtained by plotting log₁₀ y against x is log₁₀ b. Hence, for two savings models V₁ = 1000 × 1.05^t and V₂ = 1000 × 1.03^t, state each model's gradient to 4 significant figures, and explain which model has the larger percentage growth rate, without evaluating either model at any particular t.

Accepted Answer

Taking logs of y = kb^x gives log y = log k + x log b, a straight line in x with gradient log b, whatever k is. B1 for the derivation. For V₁, the gradient is log 1.05 = 0.02119; for V₂, it is log 1.03 = 0.01284, each to 4 significant figures. M1 for the method, A1 for 0.02119, A1 for 0.01284. The percentage growth rate is set by b alone, 5% for V₁ against 3% for V₂, and log is an increasing function, so the larger b gives the larger gradient: V₁ has the larger percentage growth rate. B1 for the conclusion with the reasoning.

Question 15

A radioactive isotope decays according to m = m₀e^−kt. Its mass is found to fall to 1/8 of its initial value after 30 years. Find k to 3 significant figures, and the half-life to the nearest year. Then explain how the half-life could have been read off directly from the 1/8 reading, without finding k first.

Accepted Answer

e^−30k = 1/8, so −30k = ln 1/8 = −ln 8 = −3 ln 2, and k = 3 ln 2/30 = ln 2/10 = 0.0693 per year to 3 significant figures. M1 for the equation in k, M1 for writing ln 8 as 3 ln 2, A1 for k = 0.0693. The half-life T satisfies e^−kT = ½, so T = ln 2/k = ln 2/(ln 2/10) = 10 years. M1 for T = ln 2/k, A1 for T = 10 years. Falling to 1/8 is falling to (½)³, three successive halvings, so the elapsed time of 30 years is exactly three half-lives and the half-life is 30/3 = 10 years directly, with no equation in k needed at all. B1 for that shortcut, correctly linking 1/8 to three halvings.

Question 16

A quantity y is measured at several values of x, and log₁₀ y is plotted against x, as shown in the graph. The readings are believed to fit a model of the form y = kb^x. Use the graph to find the values of k and b, giving b to 3 significant figures, and use your model to estimate y when x = 5. Explain why an estimate of y at x = 12 from the same model should be treated with caution.

Accepted Answer

Taking logarithms of y = kb^x gives log₁₀ y = log₁₀ k + x log₁₀ b, which is a straight line against x with gradient log₁₀ b and vertical intercept log₁₀ k. The line passes through the plotted points (0, 2) and (8, 6), so its gradient is (6 − 2)/(8 − 0) = 0.5. M1 for a gradient taken from two points read off the line. Then log₁₀ b = 0.5, so b = 10^0.5 = 3.16 to 3 significant figures. A1 for b. The line meets the vertical axis at 2, so log₁₀ k = 2 and k = 10² = 100. B1 for k. At x = 5 the model gives log₁₀ y = 0.5 × 5 + 2 = 4.5. M1 for substituting x = 5. So y = 10^4.5 = 3.16 × 10⁴ to 3 significant figures. A1 for the estimate. The plotted readings only cover 0 ≤ x ≤ 8, so x = 12 is well outside them: the estimate there assumes the straight line continues where nothing has been measured, and an exponential model climbs so steeply that a small error in the gradient becomes a large error in y. B1 for the caution about extrapolating beyond the data. Reading k straight off the axis as 2 mistakes the intercept for k; the intercept is log₁₀ k, not k.