Convert equation from non-linear form to linear form
Recap on logarithm laws:
Product law: $ \log_a xy = $ $ \log_a x + \log_a y $
Quotient law: $ \log_a {x \over y} = $ $ \log_a x - \log_a y $
Power law: $ \log_a x^r = $ $ r \log_a x $
Example
Convert each of the following non-linear equations, where $a$ and $b$ are constants, into the linear form $Y = mX + c$. State what the variables $X$ and $Y$ and the constants $m$ and $c$ represent.
(i) $ y = a \sqrt{x} + { b \over \sqrt{x} }$
Solutions
\begin{align*}
y & = a \sqrt{x} + {b \over \sqrt{x} } \\
\sqrt{x} (y) & = \sqrt{x} \left( a \sqrt{x} + {b \over \sqrt{x} } \right) \\
\sqrt{x} y & = a x + b \\
\\
Y & = \sqrt{x} y \\
m & = a \\
X & = x \\
c & = b
\end{align*}
(ii) $ y = {a \over x - b} $
Solutions
\begin{align*}
{y \over 1} & = {a \over x - b} \\
y(x - b) & = a \phantom{0000000000} [\text{Cross-multiply}] \\
xy - by & = a \\
xy & = by + a \\
\\
Y & = xy \\
m & = b \\
X & = y \\
c & = a
\end{align*}
(iii) $ y^b = 10^{x + a} $
Solutions
\begin{align*}
y^b & = 10^{x + a} \\
\lg y^b & = \lg 10^{x + a} \\
b \lg y & = (x + a) \lg 10 \phantom{000000} [\text{Power law}] \\
b \lg y & = (x + a) (1) \\
b \lg y & = x + a \\
\lg y & = {1 \over b} (x + a) \\
\lg y & = {1 \over b}x + {a \over b} \\
\\
Y & = \lg y \\
m & = {1 \over b} \\
X & = x \\
c & = {a \over b}
\end{align*}
(iv) $ y = ax^b + 5 $
Solutions
\begin{align*}
y & = ax^b + 5 \\
y - 5 & = ax^b \\
\ln (y - 5) & = \ln (ax^b) \phantom{0000000000} [\text{Can use } \lg \text{ as well}] \\
\ln (y - 5) & = \ln a + \ln x^b \phantom{0000000} [\text{Product law}] \\
\ln (y - 5) & = \ln a + b \ln x \phantom{000000.} [\text{Power law}] \\
\ln (y - 5) & = b \ln x + \ln a \\
\\
Y & = \ln (y - 5) \\
m & = b \\
X & = \ln x \\
c & = \ln a
\end{align*}
Questions
Find the values of constants (don't need to plot graph)
1. The diagram shows a part of a straight line to represent the curve $y = {x \over ax + b}$, where $a$ and $b$ are constants. The line passes through the points $(2, 7)$ and $(5, 1)$.
Find the value of $a$ and of $b$.
Answer: $ a = 11, b= -2 $
Solutions
\begin{align*}
{y \over 1} & = {x \over ax + b} \\
y(ax + b) & = x \phantom{000000000000} [\text{Cross-multiply}] \\
ax + b & = {x \over y} \\
{1 \over x}(ax + b) & = {1 \over y} \\
a + b\left(1 \over x\right) & = {1 \over y} \\
\\
{1 \over y} & = b \left(1 \over x\right) + a \phantom{0000} [Y = mX + c] \\
\\
\text{Gradient of line} & = { 7 - 1 \over 2 - 5 } \\
\text{Gradient of line} & = -2 \\
\implies b & = -2 \\
\\
{1 \over y} & = -2 \left(1 \over x\right) + a \\
\\
\text{Using } & (5, 1), \\
1 & = -2 (5) + a \\
1 & = -10 + a \\
11 & = a \\
\\
\therefore a & = 11, b = -2
\end{align*}
2. Variables $x$ and $y$ are related by the equation $ y = 1 + e^a x^b $, where $a$ and $b$ are constants. When a graph of $ \ln (y - 1)$ is plotted against $ \ln x $, a straight line that passes through the points $(1, 1)$ and $(2, 5)$ is obtained. Find the values of $a$ and of $b$.
Answer: $ a= - 3, b = 4 $
Solutions
\begin{align*}
y & = 1 + e^a x^b \\
y - 1 & = e^a x^b \\
\ln (y - 1) & = \ln (e^a x^b) \\
\ln (y - 1) & = \ln e^a + \ln x^b \phantom{000000} [\text{Product law}] \\
\ln (y - 1) & = a \ln e + b \ln x \phantom{00000} [\text{Power law}] \\
\ln (y - 1) & = a(1) + b \ln x \\
\ln (y - 1) & = b \ln x + a \phantom{00000000} [Y = mX + c] \\
\\
\text{Gradient of line} & = {5 - 1 \over 2 - 1} \\
& = 4 \\
\implies b & = 4 \\
\\
\ln (y - 1) & = 4 \ln x + a \\
\\
\text{Using } & (1, 1), \\
1 & = 4(1) + a \\
1 & = 4 + a \\
-3 & = a \\
\\
\therefore a & = -3, b = 4
\end{align*}
Plot graph
3. It is known that the true values of $x$ and $y$ are connected by the equation $y = ax^3 + bx^2$, where $a$ and $b$ are constants. The table gives experimental values of $x$ and $y$.
| $ x $ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
| $ y $ |
$ 10 $ |
$ 46 $ |
$ 128 $ |
$ 289 $ |
$ 442 $ |
However, one of the values of $y$ in the table was recorded wrongly.
(from A Maths 360 Ex 9.2)
(i) Plot ${y \over x^2}$ against $x$ and determine which value of $y$ is wrong.
Answer: $ y = 289 $
Solutions
| $x$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
| ${y \over x^2}$ |
$1.11$ |
$2.875$ |
$5.12$ |
$8.03$ |
$9.02$ |
\begin{align}
\text{Incorrect point: } & \left(6, 8{1 \over 36} \right) \\
\\
\therefore \text{Incorrect value of } y & = 289
\end{align}
(ii) Draw a straight line graph and estimate a value of $y$ to replace the wrong value.
Answer: $ y = 252 $
Solutions
\begin{align}
\text{Correct point: } & \left(6, 7 \right) \\
\\
{y \over x^2} & = 7 \\
{y \over 6^2} & = 7 \\
y & = 7(6^2) \\
& = 252
\end{align}
(iii) Use your graph to estimate the value of $a$ and of $b$.
Answer: $ a \approx 1.98, b = -4.9 $
Solutions
\begin{align}
y & = ax^3 + bx^2 \\
y & = x^2(ax + b) \\
\text{Linear eqn: } \phantom{000} {y \over x^2} & = ax + b \\
\\
\text{where } Y & = {y \over x^2} \\
X & = x \\
m & = a \\
c & = b \\
\\
\text{From graph, vertical intercept, } c & = -4.9 \\
\therefore b & = -4.9 \\
\\
\text{Gradient, } m & = {7 - (-2.5) \over 6 - 1.2} \\
& = 1.97916 \\
& \approx 1.98 \\
\therefore a & \approx 1.98
\end{align}
(iv) Use your graph to estimate the value of $y$ when $x = 2$.
Answer: $ y = -3.6 $
Solutions
\begin{align*}
\text{From graph, when } x & = 2, {y \over x^2} = -0.9 \\
\\
{y \over (2)^2} & = -0.9 \\
{y \over 4} & = {-0.9 \over 1} \\
y & = 4(-0.9) \\
y & = -3.6
\end{align*}
(v) By drawing a suitable line on your graph, solve the equation $0 = (a + 1)x^3 + bx^2$ for $x > 0$.
Answer: $ x = 1.6 $
Solutions
\begin{align*}
0 & = (a + 1)x^3 + bx^2 \\
0 & = ax^3 + x^3 + bx^2 \\
-x^3 & = ax^3 + bx^2 \\
-x^3 & = x^2(ax + b) \\
{-x^3 \over x^2} & = ax + b \\
-x & = \underbrace{ ax + b }_{ \text{Line drawin in ii} } \\
\\
\therefore \text{Draw } & {y \over x^2} = -x
\end{align*}
| $x$ |
$0$ |
$1$ |
$2$ |
| ${y \over x^2}$ |
$0$ |
$-1$ |
$-2$ |
\begin{align*}
\text{From graph, } x & = 1.6
\end{align*}
Real-life/geometry problem
4. A cuboid of volume $V$ cm3 has a height of $x$ cm and a rectangular base of area $(ax^2 + b)$ cm2. Corresponding values of $x$ and $V$ are shown in the table below.
| $ x $ |
$5$ |
$10$ |
$15$ |
$20$ |
| $ V $ |
$ 195 $ |
$ 840 $ |
$ 2385 $ |
$ 5280 $ |
(from think! Workbook Review Ex 8)
(i) Using suitable variables, draw a straight line graph and hence estimate the value of each of the constants $a$ and $b$.
Answer: $ a = 0.6, b = 25 $
Solutions
\begin{align}
\text{Volume of cuboid} & = \text{Base area} \times \text{Height} \\
V & = (ax^2 + b) \times x \\
V & = x(ax^2 + b) \\
{V \over x} & = ax^2 + b
\phantom{000000} [Y = mX + c] \\
\\
\text{Plot } & {V \over x} \text{ against } x^2
\end{align}
| $ x^2 $ |
$ 25 $ |
$ 100 $ |
$ 225 $ |
$ 400 $ |
| $ {V \over x} $ |
$ 39 $ |
$ 84 $ |
$ 159 $ |
$ 264 $ |
\begin{align}
\text{Vertical intercept, } c & = 25 \\
b & = 25 \\
\\
\text{Gradient, } m & = {205 - 25 \over 300 - 0} \\
a & = 0.6
\end{align}
(ii) Explain how another straight line drawn on your graph in part (i) can lead to an estimate of the value of $x$ for which the cuboid is a cube. Draw this line and find the value of $x$, correct to 1 significant figure.
Answer: $ x \approx 8 $
Solutions
\begin{align}
V & = x^3 \phantom{000000} [\text{Formula to find volume of cube}] \\
{V \over x} & = x^2
\end{align}
| $ x^2 $ |
$ 0 $ |
$ 100 $ |
$ 200 $ |
| $ {V \over x} $ |
$ 0 $ |
$ 100 $ |
$ 200 $ |
\begin{align}
\text{From graph, } x^2 & = 65 \\
x & = \pm \sqrt{65} \\
x & = \pm 8.0622 \\
\\
\therefore x & \approx 8 \text{ (1 s.f.)}
\end{align}
Past year O level questions
| Year & paper |
Comments |
| 2025 P1 Question 5 |
(a) Explanation question
(b) Find the values of constants |
| 2024 P2 Question 10 |
(a) Plot graph
(b) Explanation question |
| 2022 P1 Question 2 |
Real-life problem (Need to plot graph) |
| 2021 P2 Question 7a |
Explanation question |
| 2021 P2 Question 7b |
Real-life problem (Need to plot graph) |
| 2020 P2 Question 6 |
Real-life problem (Need to plot graph) |
| 2019 P2 Question 8 |
Real-life problem (Need to plot graph) |
| 2018 P2 Question 4 |
Real-life problem (Need to plot graph) |
| 2017 P1 Question 3 |
Find the values of constants (Can solve with or without plotting graph) |
| 2016 P2 Question 1 |
Real-life problem (Need to plot graph) |
| 2015 P2 Question 11 |
Geometry problem (Need to plot graph) |
| 2014 P1 Question 5 |
Plot graph |
| 2013 P1 Question 13 |
Real-life problem (Need to plot graph) |
| 2012 P1 Question 5 |
Real-life problem (Need to plot graph) |
| 2011 P1 Question 2 |
Find the values of constants (don't need to plot graph) |
| 2010 P1 Question 7 |
Plot graph |
| 2009 P1 Question 10 |
Real-life problem (Need to plot graph) |
| 2008 P1 Question 12 |
Graph provided (Link - Subscription required) |
| 2007 P1 Question 12 Or |
Plot graph |
| 2006 P1 Question 12 Or |
Plot graph |
| 2005 P1 Question 12 Either |
Plot graph |
| 2004 P2 Question 9 |
Linearise non-linear equations |
| 2003 P2 Question 11 |
Plot graph |
| 2002 P1 Question 12 Or |
Geometry problem (Need to plot graph) |
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