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

The areas of four lakes are 4 km², 250 km², 18 000 km² and 82 000 km².

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

   • log10 4 = 0.602
   • log10 250 = 2.40
   • log10 18000 = 4.26
   • log10 82000 = 4.91

2. (b) By how much does the log10 of an area increase when the area is multiplied by 10?

   • Multiplying by 10 adds 1 to the log.
   • Answer: 1

A rumour spreads through a school. The number of people y who have heard it after x hours follows \(y=a\times b^{x}\). The graph of \(\log_{10}y\) against x is a straight line through (0, 0.9) and (5, 2.4).

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

   • Gradient = (2.4 − 0.9) / (5 − 0) = 0.300
   • Intercept = 0.900: where the line meets the vertical axis

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

   • log b = gradient, so b = 10^0.300 = 2.00
   • log a = intercept, so a = 10^0.900 = 7.94

3. (c) Find y when x = 9.

   • Use the unrounded values: y = 10^0.900 × (10^0.300)^9 = 10^3.60
   • y = 3980

The strength y of a signal at distance x from a mast follows \(y=a\,x^{k}\). The graph of \(\log_{10}y\) against \(\log_{10}x\) is a straight line through (0, 1.3) and (2, 3.1).

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

   • Gradient = (3.1 − 1.3) / (2 − 0) = 0.900
   • Intercept = 1.30

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

   • k = gradient = 0.900
   • log a = 1.30, so a = 10^1.30 = 20.0

3. (c) Find y when x = 50.

   • Use the unrounded a: y = 10^1.30 × 50^0.900
   • y = 675

The number of people y who follow a page after x weeks is recorded. Test the model \(y=a\times b^{x}\).

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

   • log10 6.00 = 0.778
   • log10 15.0 = 1.18
   • log10 37.5 = 1.57
   • log10 93.8 = 1.97

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

   • Differences: 0.398, 0.398, 0.398
   • 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 = 0.778, so a = 6
   • Gradient log b = 0.398, so b = 2.5
   • y = 6 × 2.5^x

The number of yeast cells N (thousands) in a sample is counted every hour. 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 0.163, intercept 0.378, r = 0.9994

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

   • b = 10^0.163 = 1.45
   • a = 10^0.378 = 2.39
   • N = 2.39 × 1.45^t

3. (c) Use the model to estimate N when t = 10.

   • t = 10: N = a × b^10, using the unrounded a and b stored on the calculator
   • N = 101

The power output P (kW) of a wind turbine is measured at five wind speeds v (m/s). A model \(P=a\,v^{k}\) is suggested.

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

   • Calculate log10 v and log10 P for each row
   • Linear regression of log10 P on log10 v:
   • gradient 2.98, intercept −0.243, r = 0.9999

2. (b) Write the model in the form P = a × v^k.

   • k = gradient = 2.98
   • a = 10^−0.243 = 0.572
   • P = 0.572 × v^2.98

3. (c) Use the model to estimate P when v = 14.

   • v = 14: P = 0.572 × 14^2.98
   • P = 1490 kW

A scientist 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 = 0.9056

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 = 0.9999

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

   • The log-log graph is straighter (0.9999 against 0.9056): power
   • k = 1.60, a = 10^0.383 = 2.41
   • y = 2.41 × x^1.60

Two relationships are \(y=12\times0.8^{x}\) and \(y=5\,x^{3}\).

1. (a) Write log10 y as a linear expression in x for \(y=12\times0.8^{x}\).

   • log10 y = log10 12 + x log10 0.8
   • log10 y = 1.08 + (−0.0969)x

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

   • The gradient is log10 0.8 = −0.0969
   • The intercept is log10 12 = 1.08
   • The gradient is negative because b < 1: y is decaying

3. (c) Write log10 y as a linear expression in log10 x for \(y=5\,x^{3}\), and state its gradient and intercept.

   • log10 y = log10 5 + 3 log10 x
   • log10 y = 0.699 + 3 log10 x
   • Gradient 3, intercept 0.699