Logarithmic linearisation of a power law — 9709 P3
Read model constants from transformed axes rather than original coordinates.
Notes and key methods
For y=Axⁿ with positive x,y,A, logarithms give \(\log_b y=n\log_b x+\log_b A\). A plot of V=log_b y against U=log_b x therefore has gradient n and intercept log_b A.
For \(\displaystyle y=A b^{kx} a\) plot of log_b y against x has gradient k. Write down the transformed variables explicitly and convert an intercept back to A using the inverse exponential.
Check the conditions
Inputs and outputs of a log plot are positive.
Read the axis labels before interpreting its slope.
Three worked examples
Example 1 · Straight-line transformations of data
A positive model y=Axⁿ is plotted with \(\displaystyle U=\log _{2}x\) and \(\displaystyle V=\log _{2}y\). The line passes through \(\left( 1, \ 5\right)\) and \(\left( 3, \ 11\right)\).
Show that the stated model becomes a straight line in U and V.
Determine A and n from the transformed points.
Find A.
Find the model output y at \(\displaystyle x=2\).
Show the reasoning
(a)
Take base-2 logarithms of y=Axⁿ and use the power and product laws.
Answer: \(V=nU+\log_2 A\)
(b)
The slope is \(\displaystyle \frac{11- 5}{3- 1}= 3\). The intercept is 2, so \(\displaystyle A=2^2=4\).
Answer: \(4\) and \(3\)
(c)
Undo the base-2 logarithm of A.
\(4\)
Answer: \(4\).
(d)
Substitute into the recovered power law.
\(32\)
Answer: \(32\).
Example 2 · Straight-line transformations of data
A positive model y=Axⁿ is plotted with \(\displaystyle U=\log _{2}x\) and \(\displaystyle V=\log _{2}y\). The line passes through \(\left( 1, \ 7\right)\) and \(\left( 3, \ 15\right)\).
Show that the stated model becomes a straight line in U and V.
Determine A and n from the transformed points.
Find A.
Find the model output y at \(\displaystyle x=2\).
Show the reasoning
(a)
Take base-2 logarithms of y=Axⁿ and use the power and product laws.
Answer: \(V=nU+\log_2 A\)
(b)
The slope is \(\displaystyle \frac{15- 7}{3- 1}= 4\). The intercept is 3, so \(\displaystyle A=2^3=8\).
Answer: \(8\) and \(4\)
(c)
Undo the base-2 logarithm of A.
\(8\)
Answer: \(8\).
(d)
Substitute into the recovered power law.
\(128\)
Answer: \(128\).
Example 3 · Straight-line transformations of data
A positive model y=Axⁿ is plotted with \(\displaystyle U=\log _{2}x\) and \(\displaystyle V=\log _{2}y\). The line passes through \(\left( 1, \ 9\right)\) and \(\left( 3, \ 19\right)\).
Show that the stated model becomes a straight line in U and V.
Determine A and n from the transformed points.
Find A.
Find the model output y at \(\displaystyle x=2\).
Show the reasoning
(a)
Take base-2 logarithms of y=Axⁿ and use the power and product laws.
Answer: \(V=nU+\log_2 A\)
(b)
The slope is \(\displaystyle \frac{19- 9}{3- 1}= 5\). The intercept is 4, so \(\displaystyle A=2^4=16\).
Answer: \(16\) and \(5\)
(c)
Undo the base-2 logarithm of A.
\(16\)
Answer: \(16\).
(d)
Substitute into the recovered power law.
\(512\)
Answer: \(512\).
Common mistakes
The intercept is log_b A, not A.
A power law is not linear in x and log y unless n has a different role.