Transforming Relationships to Linear Form (Cambridge (CIE) O Level Additional Maths): Revision Note

• You may need to leave your answer in exact form, especially in the non-calculator paper

  • e.g. to solve ln a = 2 the answer will be a = e²

The heights, $h$ metres, and the amount of time spent sleeping, $t$ hours, of a group of young giraffes can be modelled using $t=a{h}^b$, where $a$ and $b$ are constants. 

The graph of $\ln t$ against $\ln h$ is a straight line passing through the points (1, -0.9) and (4, -4.5).

Find the values of $a$ and $b$, giving your answers in exact form. 

Find the gradient of the straight line between the two coordinates.

$m=\frac{-4.5--0.9}{4-1}=-\frac{3.6}{3}=-1.2$

Substitute $m=-1.2$, $x=\ln h$ and $y=\ln t$ into the equation of a straight line ($y=mx+c$).

$\ln t=-1.2\ln h+c$ 

Substitute either coordinate in and rearrange to find $c$.

$\begin{array}{rcl}-0.9& =& -1.2\left(1\right)+c\\ c& =& -0.9+1.2\\ c& =& 0.3\end{array}$

Compare this to the linearised form of $t=a{h}^b$. Take logarithms of both sides first and rearrange if you can't remember the correct form. 

$$\begin{gathered} \ln t=\ln \left(a{h}^b\right) \\ \ln t=\ln a+b\ln h \end{gathered}$$ 

So $b$ is the value in front of $\ln h$ and $\ln a=0.3$. Solve $\ln a=0.3$ by taking $\mathrm{e}$ of both sides.

$\begin{array}{rcl}\ln a& =& 0.3\\ {\mathrm{e}}^{\ln a}& =& {\mathrm{e}}^{0.3}\\ a& =& {\mathrm{e}}^{0.3}\end{array}$

$\mathit{a}\mathbf{=}{\mathbf{e}}^{\mathbf{0}\mathbf{.}\mathbf{3}}\mathbf{,}\mathit{b}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{2}$

• Unless the question specifies, you can choose whether to use ln, lg or log

• Remember you will need to solve the equation at the end

  • If using lg, solve by taking 10 to the power of each side

  • If using ln, solve by taking e to the power of each side

Variables $x$ and $y$ are such that when $\lg y$ is plotted against $x$, a straight line passing through the points $(2,5)$ and $(5,8)$ is obtained.

Show that $y=A\times b^x$ where $A$ and $b$are constants to be found. 

Find the gradient of the straight line between the two coordinates.

$m=\frac{8-5}{5-2}=\frac{3}{3}=1$

Substitute $m=1$, $X=x$ and $Y=\lg y$ into the equation of a straight line ($Y=mX+c$).

$\lg y=x+c$ 

Substitute either coordinate in and rearrange to find $c$.

$\begin{array}{rcl}5& =& 2+c\\ c& =& 3\end{array}$

$\lg y=x+3$

Linearise $y=A\times b^x$.
Take logarithms of both sides first and rearrange if you can't remember the correct form.  Take logarithms of both sides. 

$\lg y=\lg \left(A{b}^x\right)$ 

Split the right-hand side into the sum of two logarithms.

$\lg y=\lg A+\lg b^x$

Bring down the power in the final term. 

$\lg y=\lg A+x\lg b$

Compare this to the equation of the line.

$\lg b=1$ and $\lg A=3$ 

Solve $\lg b=1$ by raising 10 to the power of both sides.

$\begin{array}{rcl}\lg b& =& 1\\ {10}^{\lg b}& =& {10}^1\\ b& =& 10\end{array}$

Solve $\lg A=3$ by raising 10 to the power of both sides.

$\begin{array}{rcl}\lg A& =& 3\\ {10}^{\lg A}& =& {10}^3\\ A& =& 1000\end{array}$

$\mathit{y}\mathbf{=}\mathbf{1000}\mathbf{\times }{\mathbf{10}}^{\mathit{x}}$