Choosing Transformed Axes | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Derive the axes from the equation you are given

Do not choose axes just because a question contains an exponential. Take logs and rearrange into $Y=mX+c$ . The plotted variables must contain known expressions in the data, not unknown constants.

Worked example

Linearise y = Ae⁻ᵇˣ², where A,b > 0.

\ln y=-bx^2+\ln A

Plot $\ln y$ against $x^2$ , not $x$ . The gradient is $-b$ , so $b$ is the negative of the gradient.

Worked example

Linearise T = 20 + Ae⁻ᵇᵗ, where T > 20 and A,b > 0.

\begin{gathered}T-20=Ae^{-bt}\\\ln(T-20)\\=-bt+\ln A\end{gathered}

Subtract the known offset before taking logs. Plot $\ln(T-20)$ against $t$ . Taking $\ln T$ would not produce this straight line.

Your turnPractice

For $y=ke^{n(x-2)}$ , identify the gradient and intercept when $\ln y$ is plotted against $x$ . If they are 0.5 and 1, find $n,k$ .

Show worked answer

\ln y=nx+(\ln k-2n)

The gradient is $n$ ; the intercept is $\ln k-2n$ , not $\ln k$ .

Answer

n=0.5,\qquad k=e^2

Sometimes the original x and y are the axes

Worked example

Show that 5⁸ʸ = 6⁷ˣ gives y = kx.

This is the form in (9709/22/O/N/24 Q1). Take logs of the two positive powers.

\begin{gathered}8y\ln5=7x\ln6\\y=\frac{7\ln6}{8\ln5}x\end{gathered}

It matches $y=mx+c$ , with $m=k\approx0.974$ and $c=0$ . Plot $y$ against $x$ ; neither axis needs a logarithm.

Unknown bases can be constants in the same algebra. For $A^{2y-1}=B^{x-y}$ with $A,B>1$ , expand after taking logs:

\begin{gathered}(2y-1)\ln A=(x-y)\ln B\\y(2\ln A+\ln B)\\=x\ln B+\ln A\end{gathered}

Dividing by the positive constant $2\ln A+\ln B$ gives $y=mx+c$ . The coefficients depend only on the constants, not on $x,y$ . This is the linear-form reasoning tested in (9709/32/M/J/24 Q3).

Your turnPractice

If $A=B^2$ and $B>1$ , simplify that relationship to $y=mx+c$ .

Show worked answer

\begin{gathered}2(2y-1)=x-y\\5y=x+2\end{gathered}

Answer

y=\frac15x+\frac25

You may either substitute $\ln A=2\ln B$ in the log equation or write both powers with base $B$ .

Derive the axes from the equation you are given

Do not choose axes just because a question contains an exponential. Take logs and rearrange into $Y=mX+c$ . The plotted variables must contain known expressions in the data, not unknown constants.

Worked example

Linearise y = Ae⁻ᵇˣ², where A,b > 0.

\ln y=-bx^2+\ln A

Plot $\ln y$ against $x^2$ , not $x$ . The gradient is $-b$ , so $b$ is the negative of the gradient.

Worked example

Linearise T = 20 + Ae⁻ᵇᵗ, where T > 20 and A,b > 0.

\begin{gathered}T-20=Ae^{-bt}\\\ln(T-20)\\=-bt+\ln A\end{gathered}

Subtract the known offset before taking logs. Plot $\ln(T-20)$ against $t$ . Taking $\ln T$ would not produce this straight line.

Your turnPractice

For $y=ke^{n(x-2)}$ , identify the gradient and intercept when $\ln y$ is plotted against $x$ . If they are 0.5 and 1, find $n,k$ .

Show worked answer

\ln y=nx+(\ln k-2n)

The gradient is $n$ ; the intercept is $\ln k-2n$ , not $\ln k$ .

Answer

n=0.5,\qquad k=e^2

Sometimes the original x and y are the axes

Worked example

Show that 5⁸ʸ = 6⁷ˣ gives y = kx.

This is the form in (9709/22/O/N/24 Q1). Take logs of the two positive powers.

\begin{gathered}8y\ln5=7x\ln6\\y=\frac{7\ln6}{8\ln5}x\end{gathered}

It matches $y=mx+c$ , with $m=k\approx0.974$ and $c=0$ . Plot $y$ against $x$ ; neither axis needs a logarithm.

Unknown bases can be constants in the same algebra. For $A^{2y-1}=B^{x-y}$ with $A,B>1$ , expand after taking logs:

\begin{gathered}(2y-1)\ln A=(x-y)\ln B\\y(2\ln A+\ln B)\\=x\ln B+\ln A\end{gathered}

Dividing by the positive constant $2\ln A+\ln B$ gives $y=mx+c$ . The coefficients depend only on the constants, not on $x,y$ . This is the linear-form reasoning tested in (9709/32/M/J/24 Q3).

Your turnPractice

If $A=B^2$ and $B>1$ , simplify that relationship to $y=mx+c$ .

Show worked answer

\begin{gathered}2(2y-1)=x-y\\5y=x+2\end{gathered}

Answer

y=\frac15x+\frac25

You may either substitute $\ln A=2\ln B$ in the log equation or write both powers with base $B$ .