Revise Linearizing data: log-log and semi-log graphs for IB Mathematics: Applications and Interpretation (2.7)

This is a Paper 2 question, so the GDC does the regression; the marks are for the transform and the recovery of $k$ and $a$.\n\n**(a) Why semi-log.** Taking logs of $P = ke^{at}$ gives $\ln P = at + \ln k$, a straight line in $\ln P$ and $t$. Form the transformed table:\n\n| $t$ | $\ln P$ | \n |---|---| \n | 0 | $\ln 20 \approx 3.00$ | \n | 1 | $\ln 30 \approx 3.40$ | \n | 2 | $\ln 45 \approx 3.81$ | \n | 3 | $\ln 66 \approx 4.19$ | \n | 4 | $\ln 99 \approx 4.60$ |\n\nThe points $(t, \ln P)$ lie very close to a straight line, so the exponential model — and the semi-log graph — is appropriate. [M1: log the $P$-values] [A1: correct transformed values]\n\n**(b) Regression line.** Enter $t$ in one list and $\ln P$ in a second, and run linear regression:\n$\ln P = 0.400\,t + 3.00$ (3 s.f.). [A1][A1]\n\n**(c) Recover the parameters.** Compare $\ln P = 0.400\,t + 3.00$ with the theoretical $\ln P = a\,t + \ln k$.\nGradient: $a = 0.400$. [A1]\nIntercept: $\ln k = 3.00$, so $k = e^{3.00} = 20.0$ (3 s.f.). [M1: $k = e^{\text{intercept}}$] [A1]\n\nModel: $P = 20.0\,e^{0.400\,t}$. A quick check with the GDC's built-in exponential regression gives the same figures.

(a) Log-log regression. For a power model log BOTH variables: $\ln T = n\ln L + \ln k$. Build the transformed table:\n\n| $\ln L$ | $\ln T$ | \n |---|---| \n | $\ln 1 = 0$ | $\ln 2.0 \approx 0.693$ | \n | $\ln 2 \approx 0.693$ | $\ln 5.7 \approx 1.74$ | \n | $\ln 3 \approx 1.10$ | $\ln 10.4 \approx 2.34$ | \n | $\ln 4 \approx 1.39$ | $\ln 16.0 \approx 2.77$ | \n | $\ln 5 \approx 1.61$ | $\ln 22.4 \approx 3.11$ |\n\nEnter $\ln L$ in one list and $\ln T$ in a second and run linear regression:\n$\ln T = 1.50\,\ln L + 0.693$ (3 s.f.). [M1: log both variables] [A1][A1]\n\n**(b) Recover the parameters.** Compare with $\ln T = n\ln L + \ln k$.\nGradient: $n = 1.50$. [A1]\nIntercept: $\ln k = 0.693$, so $k = e^{0.693} = 2.00$ (3 s.f.). [M1: $k = e^{\text{intercept}}$] [A1]\n\nModel: $T = 2.00\,L^{1.50}$ — the period grows as the three-halves power of the length, matching the physics.

(a) Choose the model. The log-log correlation $r = 0.9993$ is closer to 1 than the semi-log $r = 0.918$, so the log-log transform straightens the data best. Therefore the power model $y = kx^n$ is more appropriate. [A1: compare $r$-values] [A1: correct conclusion]\n\n**(b) Build the model.** The log-log line is $\ln y = 2.10\,\ln x + 0.916$. Compare with $\ln y = n\ln x + \ln k$.\nGradient: $n = 2.10$. [A1]\nIntercept: $\ln k = 0.916$, so $k = e^{0.916} = 2.50$ (3 s.f.). [M1: $k = e^{\text{intercept}}$] [A1]\n\nModel: $y = 2.50\,x^{2.10}$.

Model answer — full working.\n\nBecause $\ln y$ is plotted against $x$, this is a semi-log graph and identifies an exponential model $y = ke^{ax}$.\n\nSet up the comparison. Taking logs of $y = ke^{ax}$ gives\n$\ln y = ax + \ln k,$\nwhich is a straight line in $\ln y$ and $x$. Compare it term by term with the given line $\ln y = 0.25x + 1.6$.\n\nGradient gives $a$. Matching the coefficients of $x$:\n$a = 0.25.$\n\nIntercept gives $k$. Matching the constant terms, $\ln k = 1.6$, so exponentiate:\n$k = e^{1.6} = 4.953\ldots \approx 4.95.$\n\nModel: $y = 4.95\,e^{0.25x}.$\n\n---\nHow our marking engine awards the 4 marks:\n\n- M1 — recognise the transform. A method mark for identifying the semi-log plot as an exponential model and setting up the comparison $\ln y = ax + \ln k$ against the given line. It is the approach that is rewarded, so it survives an arithmetic slip further down.\n- A1 — the rate $a$. Awarded for $a = 0.25$, read directly as the gradient. This accuracy mark stands on the M1 comparison above.\n- M1 — recover $k$ from the intercept. A method mark for forming $k = e^{\text{intercept}} = e^{1.6}$ — the crucial step that undoes the logarithm. The engine checks you exponentiated the intercept rather than quoting $1.6$ as $k$.\n- A1 — the value of $k$ and the model. Awarded for $k \approx 4.95$ and the assembled model $y = 4.95\,e^{0.25x}$. This A-mark depends on the M1 above and carries FT: a candidate whose intercept differed but who correctly computed $e^{\text{their intercept}}$ and wrote the matching model keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $k$ written as $e^{1.6}$, $4.95$ or $4.953$, and accepts the model with $a$ shown as $0.25$ or $\tfrac14$. Once a correct model appears, ISW (ignore subsequent working) means later restatements do not lose marks.\n\nBottom line: of the 4 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the comparison and the $k = e^{\text{intercept}}$ step are written down. A candidate who writes just '$y = 4.95e^{0.25x}$' with no comparison risks 2 method marks; a candidate who shows the transform, reads $a$ from the gradient, and exponentiates the intercept keeps the method even if the final number slips.