Using logarithmic scales on graphs Problem set B One question at a time

The masses of four animals are 0.06 g, 90 g, 7500 g and 4 800 000 g.

1. (a) Write the log10 of each value, to 3 significant figures.

   • log10 0.06 = −1.22
   • log10 90 = 1.95
   • log10 7500 = 3.88
   • log10 4800000 = 6.68

2. (b) By how much does the log10 of a mass change when the mass is divided by 100?

   • Dividing by 100 divides by 10 twice, so the log falls by 2.
   • Answer: −2

The value y (in euros) of an investment after x years follows \(y=a\times b^{x}\). The graph of \(\log_{10}y\) against x is a straight line through (0, 1.4) and (6, 2.0).

1. (a) Write down the gradient and the y-intercept of the line.

   • Gradient = (2.0 − 1.4) / (6 − 0) = 0.100
   • Intercept = 1.40: where the line meets the vertical axis

2. (b) Find the values of a and b.

   • log b = gradient, so b = 10^0.100 = 1.26
   • log a = intercept, so a = 10^1.40 = 25.1

3. (c) Find y when x = 15.

   • Use the unrounded values: y = 10^1.40 × (10^0.100)^15 = 10^2.90
   • y = 794

The surface area y of a sphere-shaped cell follows \(y=a\,x^{k}\) in terms of its volume x. The graph of \(\log_{10}y\) against \(\log_{10}x\) is a straight line through (0, 0.5) and (3, 2.3).

1. (a) Write down the gradient and the y-intercept of the line.

   • Gradient = (2.3 − 0.5) / (3 − 0) = 0.600
   • Intercept = 0.500

2. (b) Find the values of a and k.

   • k = gradient = 0.600
   • log a = 0.500, so a = 10^0.500 = 3.16

3. (c) Find y when x = 200.

   • Use the unrounded a: y = 10^0.500 × 200^0.600
   • y = 76.0

The mass y (in grams) of a growing crystal after x days is recorded. Test the model \(y=a\times b^{x}\).

1. (a) Find log10 y for each value.

   • log10 12.0 = 1.08
   • log10 19.2 = 1.28
   • log10 30.7 = 1.49
   • log10 49.2 = 1.69

2. (b) Show that the points of log y against x lie on a straight line.

   • Differences: 0.204, 0.204, 0.204
   • The differences are equal, so the semi-log graph is a straight line: exponential

3. (c) Write the model, in the form a × b^x.

   • Intercept log a = 1.08, so a = 12
   • Gradient log b = 0.204, so b = 1.6
   • y = 12 × 1.6^x

The activity N (counts per minute) of a radioactive sample is measured every day. A model \(N=a\times b^{t}\) is suggested.

1. (a) Use your calculator to fit a straight line to log10 N against t. Write down the gradient, the intercept and r.

   • Calculate log10 N for each t
   • Linear regression of log10 N on t:
   • gradient −0.0861, intercept 2.98, r = −0.9986

2. (b) Write the model in the form N = a × b^t.

   • b = 10^−0.0861 = 0.820
   • a = 10^2.98 = 952
   • N = 952 × 0.820^t

3. (c) Use the model to estimate N when t = 15.

   • t = 15: N = 952 × 0.820^15
   • N = 48.7

The time T (s) for a pendulum to swing back and forth is measured for five lengths L (m). A model \(T=a\,L^{k}\) is suggested.

1. (a) Use your calculator to fit a straight line to log10 T against log10 L. Write down the gradient, the intercept and r.

   • Calculate log10 L and log10 T for each row
   • Linear regression of log10 T on log10 L:
   • gradient 0.498, intercept 0.302, r = 0.9990

2. (b) Write the model in the form T = a × L^k.

   • k = gradient = 0.498
   • a = 10^0.302 = 2.00
   • T = 2.00 × L^0.498

3. (c) Use the model to estimate T when L = 10.

   • L = 10: T = 2.00 × 10^0.498
   • T = 6.31 s

An engineer records y for six values of x and must choose between an exponential model and a power model.

1. (a) Find the correlation coefficient r for log10 y against x.

   • Calculate log10 y for each x
   • Linear regression of log10 y on x: r = 0.9998

2. (b) Find the correlation coefficient r for log10 y against log10 x.

   • Calculate log10 x for each x
   • Linear regression of log10 y on log10 x: r = 0.9664

3. (c) State which model fits best and write it in the form y = a × b^x.

   • The semi-log graph is straighter (0.9998 against 0.9664): exponential
   • b = 10^0.278 = 1.90, a = 10^0.702 = 5.04
   • y = 5.04 × 1.90^x

Two relationships are \(y=40\times0.65^{x}\) and \(y=8\,x^{2}\).

1. (a) Write log10 y as a linear expression in x for \(y=40\times0.65^{x}\).

   • log10 y = log10 40 + x log10 0.65
   • log10 y = 1.60 + (−0.187)x

2. (b) State the gradient and the intercept of that line, and say why the gradient is negative.

   • The gradient is log10 0.65 = −0.187
   • The intercept is log10 40 = 1.60
   • The gradient is negative because b < 1: y is decaying

3. (c) Write log10 y as a linear expression in log10 x for \(y=8\,x^{2}\), and state its gradient and intercept.

   • log10 y = log10 8 + 2 log10 x
   • log10 y = 0.903 + 2 log10 x
   • Gradient 2, intercept 0.903