Linearising Power Models | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Transform the variables, not the constants

Linearising means changing what is plotted so that a relationship becomes a straight line. For $y=kx^n$ with $x,k>0$ , take logs:

\begin{gathered}\ln y=\ln(kx^n)\\=n\ln x+\ln k.\end{gathered}

In $Y=mX+c$ , the gradient $m$ is the vertical change per unit horizontal change. The vertical intercept $c$ is the output when $X=0$ .

Compare this with $Y=mX+c$ . Capital $X,Y$ name the new plotted coordinates: $X=\ln x$ horizontally and $Y=\ln y$ vertically. The gradient is $m=n$ , and the vertical intercept is $c=\ln k$ .

The original point $(1,k)$ becomes $(0,\ln k)$ because $\ln1=0$ . The log graph’s intercept does not mean the original $x$ is zero.

Change $n$ , then $k$ . Predict which changes the straight line’s gradient and which changes its intercept.

Power n = 1.5

Scale k = 2

Gradient = 1.5; intercept = ln k = 0.69.

Both plots show 0.5 ≤ x ≤ 4. The green point (1,k) becomes (0,ln k). The original vertical scale adjusts; the log scale stays fixed.

Your turnPractice

For $y=5x^{-2}$ , state the axes that give a straight line, its gradient and its vertical intercept.

Show worked answer

\ln y=-2\ln x+\ln5

Plot $\ln y$ against $\ln x$ . Gradient −2; intercept $\ln5$ , not 5.

Read the line, then recover k

A plot “of $Y$ against $X$ ” puts $Y$ vertically. The gradient is vertical change divided by horizontal change:

m=\frac{Y_2-Y_1}{X_2-X_1}

A graph of ln y against ln x passes through (0,2) and (2,−1). Find $y=kx^n$ .

Use the plotted coordinates: ΔY / ΔX = −3 / 2.

n=\frac{-1-2}{2-0}=-\frac32

The point at $X=0$ gives $\ln k=2$ , so $k=e^2$ .

Answer

y=e^2x^{-3/2}

Check the second transformed point: $-3(2)/2+2=-1$ . Its original coordinates are $(e^2,e^{-1})$ , not (2,−1).

Your turnPractice

A plot of $\ln y$ against $\ln x$ passes through (1,5) and (3,9). Find $y=kx^n$ exactly.

Show worked answer

\begin{gathered}n=\frac{9-5}{3-1}=2\\5=2(1)+c\\c=3\end{gathered}

Answer

y=e^3x^2

$c=\ln k$ , so $k=e^3$ .

Transform the variables, not the constants

Linearising means changing what is plotted so that a relationship becomes a straight line. For $y=kx^n$ with $x,k>0$ , take logs:

\begin{gathered}\ln y=\ln(kx^n)\\=n\ln x+\ln k.\end{gathered}

In $Y=mX+c$ , the gradient $m$ is the vertical change per unit horizontal change. The vertical intercept $c$ is the output when $X=0$ .

Compare this with $Y=mX+c$ . Capital $X,Y$ name the new plotted coordinates: $X=\ln x$ horizontally and $Y=\ln y$ vertically. The gradient is $m=n$ , and the vertical intercept is $c=\ln k$ .

The original point $(1,k)$ becomes $(0,\ln k)$ because $\ln1=0$ . The log graph’s intercept does not mean the original $x$ is zero.

Change $n$ , then $k$ . Predict which changes the straight line’s gradient and which changes its intercept.

Power n = 1.5

Scale k = 2

Gradient = 1.5; intercept = ln k = 0.69.

Both plots show 0.5 ≤ x ≤ 4. The green point (1,k) becomes (0,ln k). The original vertical scale adjusts; the log scale stays fixed.

Your turnPractice

For $y=5x^{-2}$ , state the axes that give a straight line, its gradient and its vertical intercept.

Show worked answer

\ln y=-2\ln x+\ln5

Plot $\ln y$ against $\ln x$ . Gradient −2; intercept $\ln5$ , not 5.

Read the line, then recover k

A plot “of $Y$ against $X$ ” puts $Y$ vertically. The gradient is vertical change divided by horizontal change:

m=\frac{Y_2-Y_1}{X_2-X_1}

A graph of ln y against ln x passes through (0,2) and (2,−1). Find $y=kx^n$ .

Use the plotted coordinates: ΔY / ΔX = −3 / 2.

n=\frac{-1-2}{2-0}=-\frac32

The point at $X=0$ gives $\ln k=2$ , so $k=e^2$ .

Answer

y=e^2x^{-3/2}

Check the second transformed point: $-3(2)/2+2=-1$ . Its original coordinates are $(e^2,e^{-1})$ , not (2,−1).

Your turnPractice

A plot of $\ln y$ against $\ln x$ passes through (1,5) and (3,9). Find $y=kx^n$ exactly.

Show worked answer

\begin{gathered}n=\frac{9-5}{3-1}=2\\5=2(1)+c\\c=3\end{gathered}

Answer

y=e^3x^2

$c=\ln k$ , so $k=e^3$ .