Linearising Exponential Models | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

An exponential model needs only one logged axis

For $y=ka^x$ , where $k,a>0$ , the power law puts $x$ outside the log:

\ln y=x\ln a+\ln k

Compare with $Y=mX+c$ : plot $Y=\ln y$ against $X=x$ . The gradient is $m=\ln a$ and the vertical intercept is $c=\ln k$ . Recover the constants using $a=e^m$ and $k=e^c$ .

For $y=ke^{bx}$ , the same axes give $\ln y=bx+\ln k$ . Here the gradient is $b$ itself, because $\ln e=1$ .

Worked example

ln y against x passes through (1,1.6) and (5,4.0). Find y = kaˣ.

\begin{gathered}m=\frac{4.0-1.6}{5-1}=0.6\\1.6=0.6(1)+c\\c=1\\a=e^{0.6},\qquad k=e\end{gathered}

Answer

y=e(e^{0.6})^x=e^{1+0.6x}

Thus $a\approx1.82$ and $k\approx2.72$ . Check the other point: $0.6(5)+1=4$ .

Your turnPractice

A plot of $\ln y$ against $x$ has gradient −0.4 and intercept 2. Find $y=Ae^{bx}$ . Is it growth or decay?

Show worked answer

Answer

y=e^2e^{-0.4x}

$A=e^2$ and $b=-0.4$ . It is decay.

Use the inverse of the log base on the graph

With base-10 axes, the intercept is $\log_{10}k$ , not $\ln k$ . Undo it with $10^x$ , not $e^x$ .

Worked example

A graph of log₁₀ y against x has equation Y = 0.5X + 2.

\begin{gathered}\log_{10}y=0.5x+2\\y=10^{0.5x+2}\\y=100(10^{0.5})^x\end{gathered}

For $y=ka^x$ , this gives $k=100$ and $a=\sqrt{10}$ .

Your turnPractice

A plot of $\log_{10}y$ against $\log_{10}x$ has gradient 3 and intercept −1. Find $y$ in terms of $x$ .

Show worked answer

Answer

y=0.1x^3

The two logged axes identify a power model. Its exponent is 3 and its scale is $10^{-1}$ .

An exponential model needs only one logged axis

For $y=ka^x$ , where $k,a>0$ , the power law puts $x$ outside the log:

\ln y=x\ln a+\ln k

Compare with $Y=mX+c$ : plot $Y=\ln y$ against $X=x$ . The gradient is $m=\ln a$ and the vertical intercept is $c=\ln k$ . Recover the constants using $a=e^m$ and $k=e^c$ .

For $y=ke^{bx}$ , the same axes give $\ln y=bx+\ln k$ . Here the gradient is $b$ itself, because $\ln e=1$ .

Worked example

ln y against x passes through (1,1.6) and (5,4.0). Find y = kaˣ.

\begin{gathered}m=\frac{4.0-1.6}{5-1}=0.6\\1.6=0.6(1)+c\\c=1\\a=e^{0.6},\qquad k=e\end{gathered}

Answer

y=e(e^{0.6})^x=e^{1+0.6x}

Thus $a\approx1.82$ and $k\approx2.72$ . Check the other point: $0.6(5)+1=4$ .

Your turnPractice

A plot of $\ln y$ against $x$ has gradient −0.4 and intercept 2. Find $y=Ae^{bx}$ . Is it growth or decay?

Show worked answer

Answer

y=e^2e^{-0.4x}

$A=e^2$ and $b=-0.4$ . It is decay.

Use the inverse of the log base on the graph

With base-10 axes, the intercept is $\log_{10}k$ , not $\ln k$ . Undo it with $10^x$ , not $e^x$ .

Worked example

A graph of log₁₀ y against x has equation Y = 0.5X + 2.

\begin{gathered}\log_{10}y=0.5x+2\\y=10^{0.5x+2}\\y=100(10^{0.5})^x\end{gathered}

For $y=ka^x$ , this gives $k=100$ and $a=\sqrt{10}$ .

Your turnPractice

A plot of $\log_{10}y$ against $\log_{10}x$ has gradient 3 and intercept −1. Find $y$ in terms of $x$ .

Show worked answer

Answer

y=0.1x^3

The two logged axes identify a power model. Its exponent is 3 and its scale is $10^{-1}$ .