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Log graphs and exponential models questions
Data does not arrive labelled with its own formula. Plotting log y against log x turns a power law into a straight line, and plotting log y against x does the same for an exponential, with the gradient and intercept giving the constants. The lesson then covers using such a model and stating where it stops being reasonable.
7 original questions · 25 marks · the log graphs and exponential models notes · Exponentials and logarithms
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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.
Worked 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, and reporting it as a is where this loses a mark.A quantity is believed to follow exponential growth y = abx. State what should be plotted, and what the gradient and intercept give.
Worked 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. Which axis gets logged is the only difference between the two tests, and it is decided by where the variable sits. In y = abx the x is an exponent, so it is already the thing a log would produce and needs no logging.A quantity follows y = axn, and measurements give (4, 40) and (9, 135). Find n and a, and state the law.
Worked 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 = 2.251.5. Then a = 40/41.5 = 40/8 = 5, and the law is y = 5x1.5. 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 × 91.5 = 5 × 27 = 135.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.
Worked 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; carrying it through to write ln 30 is the usual wrong turn.The model y = 3 × 5x 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.
Worked 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. The growth factor lives in the gradient, the starting value in the intercept.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.
Worked answer
The line says log y = 0.5 log x + log 6, so the law is y = 6x0.5 = 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. Undoing the logs turns the intercept back into the front constant and the gradient back into the power.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.
Worked 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. Equal time steps multiply the mass by equal factors: the signature no other kind of curve shares.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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