Linearizing data with logs; log-log and semi-log graphs practice questions.

A data set is modelled by y = a·x^n (a power model). To linearise it, you plot ln y against ln x. What does the gradient of this line represent?

• ln a
• the value of a
• the exponent n
• ln n

Answer: the exponent n

Taking logs of y = a·x^n gives ln y = ln a + n·ln x, so plotting ln y against ln x is a straight line of gradient n and intercept ln a.

An exponential model y = a·b^x is linearised by plotting log y against x. State what the vertical-axis intercept of the resulting straight line equals.

• log a
• log b
• a
• b

Answer: log a

log y = log a + x·log b, so the line has intercept log a (when x = 0) and gradient log b.

A semi-log plot (log₁₀ y against x) of some data gives a straight line with gradient 0.30 and intercept 1. The model is y = a·b^x. Determine the value of b.

• b = 0.30
• b = 2
• b = 10
• b = 1.35

Answer: b = 2

The gradient equals log b, so b = 10^0.30 ≈ 2.0. (The intercept 1 gives a = 10^1 = 10.)

For the linearised power model log y = 0.7 + 2.5·log x, calculate the predicted value of y when x = 4, to 3 s.f.

• 160
• 80.2
• 12.6
• 5.01

Answer: 160

y = a·x^n with a = 10^0.7 = 5.012 and n = 2.5, so y = 5.012 × 4^2.5 = 5.012 × 32 = 160 (3 s.f.).

A student plots ln y against x and obtains a straight line, then plots ln y against ln x and obtains a curve. Determine which model best fits the data.

• Power model y = a·x^n
• Exponential model y = a·b^x
• Linear model y = mx + c
• Quadratic model y = ax² + bx + c

Answer: Exponential model y = a·b^x

A straight line on semi-log axes (ln y vs x) indicates an exponential model; a power model would instead be straight on log-log axes.

Taking natural logs of the power model y = a·x^n produces which linear relationship?

• ln y = ln a + n·ln x
• ln y = n·ln a + ln x
• ln y = a + n·x
• ln y = ln a + ln n · x

Answer: ln y = ln a + n·ln x

ln(a·x^n) = ln a + ln(x^n) = ln a + n·ln x, a straight line of ln y against ln x.