Logo
Δβα\Delta_{\beta \alpha}
Light
/
Dark
Zoom in
/
Zoom out
[
Δβα\Delta_{\beta \alpha}
/1]
Surds and Logarithms
[
Δβα\Delta_{\beta \alpha}
/
1.0]
Manipulating Surds
[
Δβα\Delta_{\beta \alpha}
/
1.1]
Graphs of Surd Functions
[
Δβα\Delta_{\beta \alpha}
/
1.2]
Introduction to Logarithms
[
Δβα\Delta_{\beta \alpha}
/
1.3]
Further Logarithms
[
Δβα\Delta_{\beta \alpha}
/
1.4]
Eulers Number
[
Δβα\Delta_{\beta \alpha}
/
1.ECQ]
End of Chapter Questions
Check out past Pure Mathematics papers here!
Markschemes and examiner reports included
This project was created and is maintained by 9662e103-129a (View changelog here)
1.0 Manipulating Surds
Surds are
irrational numbers
that cannot be expressed as a fraction of two integers. They are represented by the form
x\sqrt{x}
(or
x12x^{\frac{1}{2}}
in index form) and can be manipulated in the following ways:
Note 1.0.0 - Surd manipulation
1) Multiplying surds:
a×b=ab\sqrt{a}\times\sqrt{b}=\sqrt{ab}


2) Multiplying similar surds:
a×a=a\sqrt{a}\times\sqrt{a}=a


3) Dividing surds:
bb=ab\frac{\sqrt{b}}{\sqrt{b}}=\sqrt{\frac{a}{b}}


4) Rationalising surds:
ab+c×a(b−c)b−c=ab−acb−c\frac{a}{\sqrt{b}+\sqrt{c}}\times\frac{a\left( \sqrt{b}-\sqrt{c} \right)}{\sqrt{b}-\sqrt{c}}=\frac{a\sqrt{b}-a\sqrt{c}}{b-c}
Surds are generally used in precision applications like reducing noise to detect patterns in signals in computer science, all they way to our current understanding of the underlying mathematics governing quantum mechanics (such as the logic behind the Hadamard gate). Some examples to get you started are presented below, and a series of questions follow that.
Example 1.0.0 - Basic surd simplification and rationalization
1) Simplify
12\sqrt{12}


2) Simplify
202\frac{\sqrt{20}}{2}


3) Rationalise
26−2\frac{2}{\sqrt{6}-2}
giving your answer in the form
a+b6a+b\sqrt{6}
1) When simplifying surds, we need to see what can be factored out of the number under the square root. In this case,
12\sqrt{12}
. We can see that 12 can be factored into 4 and 3, and since 4 is a perfect square, we can factor it out of the square root (as 2). This gives us
4×3=4×3=23\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}=2\sqrt{3}
.

2) When simplifying fractions with surds, we need to
rationalise
the denominator. In this case, rationalizing is not needed because we can easily factor 20 out into 4 and 5, and since 4 is a perfect square, we can factor it out of the square root. This gives us
252=5\frac{2\sqrt{5}}{2}=\sqrt{5}
.

3) When rationalizing, we need to multiply the numerator and denominator by the conjugate of the denominator. In this case, the conjugate of
6−2\sqrt{6}-2
is
6+2\sqrt{6}+2
, so we multiply the numerator and denominator by this conjugate. This gives us
2(6+2)6−4=26+42=2+6\frac{2\left( \sqrt{6}+2 \right)}{6-4}=\frac{2\sqrt{6}+4}{2}=2+\sqrt{6}
.
The examples above were just simple representations, but they provide the axioms for which you can solve more complex surd problems. The questions below will test your understanding of basic surd manipulation, and further questions involving rationalization. Try to do these without a calculator!
Problem 1.0.0 - Surd manipulation and rationalization
1) Simplify
18\sqrt{18}


2) Simplify
453\frac{\sqrt{45}}{3}


3) Simplify
23×562\sqrt{3}\times 5\sqrt{6}


4) Simplify
123\frac{\sqrt{12}}{\sqrt{3}}


5) Simplify
155\frac{\sqrt{15}}{\sqrt{5}}


6) Simplify
50−18\sqrt{50}-\sqrt{18}
, giving your answer in the form
a2a\sqrt{2}


7) Simplify
500+125\sqrt{500}+\sqrt{125}
, giving your answer in the form
a5a\sqrt{5}


8) Simplify
298+502\sqrt{98}+\sqrt{50}


9) Simplify
972+689\sqrt{72}+6\sqrt{8}


10) Rationalise
5x2−2x\frac{5x}{\sqrt{2}-2x}
Problem 1.0.1 - Further rationalization
1) Rationalise
3x2−4xx+2\frac{3x^2-4x}{\sqrt{x}+2}


2) Express
2+77−2\frac{2+\sqrt{7}}{\sqrt{7}-2}
in the form
a+b7a+b\sqrt{7}


3) Express
264−3\frac{26}{4-\sqrt{3}}
in the form
a+b3a+b\sqrt{3}


4) Show that
218−22−1\frac{2\sqrt{18}-2}{\sqrt{2}-1}
can be written in the form
a+b2a+b\sqrt{2}


5) Rationalise
5+105−25\frac{\sqrt{5}+10}{5-2\sqrt{5}}
, giving your answer in the form
a+b5a+b\sqrt{5}


6) Show that
(8+2)232−4\frac{(\sqrt{8}+\sqrt{2})^2}{3\sqrt{2}-4}
can be written in the form
a+b2a+b\sqrt{2}


7) Show that
3−125(2+5)2\frac{3-\sqrt{125}}{(2+\sqrt{5})^2}
can be written in the form
a+b5a+b\sqrt{5}


8) Show that
(6+6)23+6\frac{(\sqrt{6}+6)^2}{3+\sqrt{6}}
can be written in the form
a+b6a+b\sqrt{6}


9) Solve
75z+12=3z\sqrt{75}z+12=\sqrt{3}z
, giving you answer in the form
a3a\sqrt{3}


10) Solve
3x=6x+93x=\sqrt{6}x+9
, giving your answer in the form
a+b6a+b\sqrt{6}
1.1 Graphs of Surd Functions
When trying to solve surds involving unknown variables (e.g. when trying to find the x-intercept of a surd graph), you might have noticed that it's very hard to get whole (or close to whole) number answers. This further proves how you're not able to express surds as simple fractions! If we look at the example below of the function
7x2−3\frac{\sqrt{7x}}{2}-3
, we can find that the x-values end up being large decimals.
Graph 1.1.0 - Graph of
7x2−3\frac{\sqrt{7x}}{2}-3
The graph of a surd function
7x2−3\frac{\sqrt{7x}}{2}-3
is a curve that starts at the origin and moves in a positive direction along the x-axis, with an initial but sudden increase of y values near the origin. We can apply the shape of this graph in real life to the velocity of a pendulum. As you let go of the pendulum at a maximum point, its velocity is sudden and large, but as it gets closer to the point of equilibrium, its velocity slowes down (hence the smoother gradient change in later x values).
1.2 Introduction to Logarithms
Logarithms are the inverse of exponentials, and are used to solve equations where the variable is in the exponent. The are used when dealing with real-life problems like half-life decay graphs for radioisotopes, the decibel scale, the pH scale, and the Richter scale. They are generally expressed in the form
log⁡xa\log_xa
, but can be expressed in terms of
ee
(which I will cover later in this chapter). Below are the general rules involving log conversions:
Note 1.2.0 - Converting to log form
1) General log:
N=ax→log⁡aN=xN=a^x \rightarrow \log_aN=x


2) Common log:
N=10x→log⁡N=xN=10^x \rightarrow \log N=x


3) Natural log:
N=ex→ln⁡N=xN=e^x \rightarrow \ln N=x


You may have noticed, there are several ways to convert TO a logarithm. Remember, a logarithm is the inverse of an exponential, and if we try to solve it using regular algebra rules, we find ourselfs unable to. This is where converting to log form can help especially when finding variables WITHIN exponents. Below are some examples to get you started, followed by some simple conversion questions.
Example 1.2.0 - Basic log conversion
1) Convert
25=322^5=32
to log form

2) Convert
102=10010^2=100
to log form

3) Convert
e3=20e^3=20
to log form
1) To convert
25=322^5=32
to log form, we need to see what the base is. In this case, the base is 2, and the exponent is 5. This gives us
log⁡232=5\log_232=5


2) To convert
102=10010^2=100
to log form, we need to see what the base is. In this case, the base is 10, and the exponent is 2. This gives us
log⁡100=2\log100=2


3) To convert
e3=20e^3=20
to log form, we need to see what the base is. In this case, the base is
ee
, and the exponent is 3. This gives us
ln⁡20=3\ln20=3
Above show simple conversions, but now we need to test them out in functions. Below are some examples.
Problem 1.2.0 - Log usage
1) Solve
log⁡3x=4\log_3x=4


2) Solve
log⁡x243=5\log_x243=5


3) Solve
ln⁡x=2\ln x=2


4) Solve
log⁡2x=3\log_2x=3


5) Solve
log⁡5x=1\log_5x=1


6) Solve
ln⁡x=3\ln x=3


7) Solve
log⁡4x=2\log_4x=2


8) Solve
log⁡7x=3\log_7x=3


9) Solve
ln⁡x=4\ln x=4


10) Solve
log⁡6x=2\log_6x=2
Now that you're familiar with basic log conversion, here's some formulae you'll need to memorize to be able to solve more complex log equations.
Note 1.2.1 - Log rules
1) Multiplying logs:
log⁡aN+log⁡aM=log⁡aNM\log_aN+\log_aM=\log_aNM


2) Dividing logs:
log⁡aN−log⁡aM=log⁡aNM\log_aN-\log_aM=\log_a\frac{N}{M}


3) Power conversion:
log⁡aNm=mlog⁡aN\log_aN^m=m\log_aN


4) Base conversion:
log⁡aN=log⁡bNlog⁡ba\log_aN=\frac{\log_bN}{\log_ba}


5) Equality: when
log⁡bn=log⁡bm\log_bn=\log_bm
, then
n=mn=m
Below are some examples of the application of these rules. For now, you wont need to know the 4th conversion, as it will be covered in the next section.
Example 1.2.1 - Solving equations involving logs
1) Solve
2log⁡10x=1+log⁡105−log⁡1022\log_{10}x=1+\log_{10}5-\log_{10}2


2) Solve
log⁡6x+log⁡6(x+5)=2\log_6x+\log_6(x+5)=2
, where
x>0x>0


3) Solve
log⁡3(2x−1)=2\log_3(2x-1)=2
1) To solve
2log⁡10x=1+log⁡105−log⁡1022\log_{10}x=1+\log_{10}5-\log_{10}2
, we first move all the logs to one side, and then use both the multiplication and division rule like so: multiply
log⁡10x2\log_{10}x^2
and
log⁡102\log_{10}2
and put it all over 5. This gives us
log⁡10(2x25)=1\log_{10} \left( \frac{2x^2}{5} \right)=1
. We can then remove the log, and solve like a regular equation:
2x25=10\frac{2x^2}{5}=10
, then
x2=25x^2=25
, and finally
x=5x=5
.

2) To solve
log⁡6x+log⁡6(x+5)=2\log_6x+\log_6(x+5)=2
, we can use the multiplication rule to combine the logs into one. This gives us
log⁡6(x(x+5))=2\log_6(x(x+5))=2
, and we can then remove the log to get
x(x+5)=36x(x+5)=36
, which gives us
x=4x=4
(you can either complete the square, or use the quadratic formula to get this answer).

3) To solve
log⁡3(2x−1)=2\log_3(2x-1)=2
, we can remove the log to get
2x−1=92x-1=9
, and then solve for x to get
x=5x=5
.
Now that you've seen some examples, here are some problems for you to try out.
Problem 1.2.1 - Solving and manipulating logs
1) Solve
log⁡4x+log⁡4(x+1)=1\log_4x+\log_4(x+1)=1


2) Given that
2log⁡4(3x+5)=log⁡4(3x+8)+1,x>−532\log_4(3x+5)=\log_4(3x+8)+1,x>-\frac{5}{3}
, show that
9x2+ax+b=09x^2+ax+b=0
where
aa
and
bb
are constants to be found

3) Solve
7x+2=37^{x+2}=3
, giving your answer in the form
x=log⁡7ax=\log_7a
where
aa
is a rational number in its simplest form

4) Solve
log⁡5(3−2x)+log⁡5(2+x)=1\log_5(3-2x)+\log_5(2+x)=1
where
xx
is a real number

5) Express
2log⁡3x−log⁡3(x+4)2\log_3x-\log_3(x+4)
as a single logarithm

6) Find any real values of
xx
such that
2log⁡4(2−x)−log⁡4(x+5)=12\log_4(2-x)-\log_4(x+5)=1


7) Solve the equation
log⁡2x+2log⁡23=log⁡2(x+5)\log_2x+2\log_23=\log_2(x+5)


8) Given that
3log⁡3(2x−1)=2+log⁡3(14x−25)3\log_3(2x-1)=2+\log_3(14x-25)
, show that
ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0
where
a,b,ca, b, c
and
dd
are integers to be found

9) Find the exact solution of
log⁡4(3y+2)=log⁡4(y+2)+log⁡4(2y−3)\log_4(3y+2)=\log_4(y+2)+\log_4(2y-3)


10) Solve
2log⁡2(x+1)=log⁡2(x+13)+12\log_2(x+1)=\log_2(x+13)+1
1.3 Further Logarithms
Now that you know the basic rules of log manipulation and can solve them, we need to understand how base changes work and how they can help us. The questions above all had the same base, which makes them easy to solve, but there are log functions that require base changes.
Example 1.3.0 - Base changes
1) Changing bases:
log⁡aN=log⁡bNlog⁡ba\log_aN=\frac{\log_bN}{\log_ba}
If we look above, we see that another base 'b' is created. We can derive this from other log rules as followed:
1) Let:
log⁡aN=x\log_aN=x

2) Convert to exponential form:
N=axN=a^x

3) Take the log (
log⁡b\log_b
) of the new bases:
log⁡bN=log⁡bax\log_bN=\log_ba^x

4) Use the power rule:
log⁡bN=xlog⁡ba\log_bN=x\log_ba

5) Divide by
log⁡ba\log_ba
:
log⁡aN=log⁡bNlog⁡ba\log_aN=\frac{\log_bN}{\log_ba}
Now lets take a look at some examples. These are all completed using base change rules, and when completing the questions below, you should also use base changes!
Example 1.3.1 - Simplifying equations using base changes
1) Find the value of
log⁡918\log_918


2) Find the value of
xx
in
3x=233^x=23


3) Given that
log⁡23=log⁡43x\log_23=\log_43^x
, find the value of
xx


1) To find the value of
log⁡918\log_918
, we need to change the base. We can change it to use base 10 (remember
log⁡10x=log⁡x\log_{10}x=\log x
) which gives us
log⁡18log⁡9=1.32\frac{\log 18}{\log 9}=1.32
.

2) We first need to convert to exponential form which would give
log⁡323=x\log_323=x
. From there, we apply the base change rule to get
log⁡23log⁡3=2.85\frac{\log 23}{\log 3}=2.85
.

3) We don't need to use the power rule at the start here, so we will focus yet again on changing the base. We will apply this rule to both sides to get
log⁡23log⁡22=log⁡23xlog⁡24\frac{\log_23}{\log_22}=\frac{\log_23^x}{\log_24}
. We then simplify the logs which gets us
log⁡23=log⁡23x2\log_23=\frac{\log_23^x}{2}
. We can finally use the power rule here to show that
2log⁡23=log⁡23x∴x=22\log_23=\log_23^x \therefore x=2


The examples above show how base changes can be used to simplify log equations. The questions below will test your understanding of base changes and how they can be used to simplify log equations. They won't be hard because you won't find many questions about base changes.
Problem 1.3.0 - Further log questions involving base changes
1) Find the value of
log⁡512\log_512


2) Find the value of
xx
in
2x=322^x=32


3) Given that
log⁡32=log⁡22x\log_32=\log_22^x
, find the value of
xx


4) Find the value of
log⁡218\log_2\frac{1}{8}


5) Find the value of
xx
in
5x=1255^x=125


6) Given that
log⁡125=log⁡55x\log_125=\log_55^x
, find the value of
xx


7) Find the value of
log⁡3127\log_3\frac{1}{27}


8) Find the value of
xx
in
4x=644^x=64


9) Given that
log⁡64=log⁡44x\log_64=\log_44^x
, find the value of
xx


10) Find the value of
log⁡4164\log_4\frac{1}{64}
1.4 Eulers Number
Eulers number
ee
is a mathematical constant (just like
π\pi
) which can be used to explain things from population models all the way to radioactive decay. As you saw earlier, it has the definition
N=ex→ln⁡N=xN=e^x \rightarrow \ln N=x
. Below are some examples to get you started!
Example 1.4.0 - Solving equations with
ee
1) Find the exact solution, in their simplest form, to the equation
e3x−9=8e^{3x-9}=8


2) Solve for
xx
:
5e4x−3=115e^{4x-3}=11
, giving your solution in exact form

3) Solve for
xx
:
ln⁡(3x−7)+5=8\ln (3x-7)+5=8
, giving your solution in exact form
1) We first have to convert the equation to a logarithmic form, which gives us
ln⁡8=3x−9\ln 8=3x-9
. We can then solve for
xx
to get
x=9+ln⁡83x=\frac{9+\ln 8}{3}
. This can be simplified to
ln⁡2+3\ln 2+3
.

2) We first have to move the constant to the other side so we're able to log the equation:
e4x−3=115e^{4x-3}=\frac{11}{5}
, and then
4x−3=ln⁡(115)4x-3=\ln \left( \frac{11}{5}\right)
. Then, same as always, we just rearrange to get the exact value:
4x=ln⁡(115)+34x=\ln \left( \frac{11}{5}\right)+3
, which results in:
x=ln⁡(115)+34x=\frac{\ln \left( \frac{11}{5}\right)+3}{4}
.

3) As per usual, we have to rearrange the equation to get
ln⁡(3x−7)=3\ln (3x-7)=3
, and then
3x−7=e33x-7=e^3
, and finally
x=e3+73x=\frac{e^3+7}{3}
.
There's not much else for now, so here's a few questions on solving logs with eulers number:
Problem 1.4.0 - Solving equations with
ee
1) The curve C has parametric equations
x=ln⁡tx=\ln t
and
y=t2−2y=t^2-2
when
t>0t>0
. Find the cartesian equation of C

2) The value of a car can be calculated from the formula
V=17000e−0.25t+2000e−0.5t+500V=17000e^{-0.25t}+2000e^{-0.5t}+500
where
VV
is the value of the car in dollars and
tt
is the age in years. Calculate the exact value of
tt
when
V=9500V=9500


3) The population
pp
of a species at time
tt
years is mapped by the equation
p=2800ae0.2t1+ae0.2tp=\frac{2800ae^{0.2t}}{1+ae^{0.2t}}
where
aa
is a constant. Use the equation with
a=0.12a=0.12
to predict the number of years before the population of the species reaches 1850

4) A curve with the equation
(8−x)ln⁡x(8-x)\ln x
(where
x>0x>0
) cuts the x-axis at two points. What are the
xx
-coordinates of these points?

5) Without using a calculator, solve an equation to find the exact
xx
-coordinate of the intersection points of the functions
y=e2xy=e^{2x}
and
y=8e−xy=8e^{-x}


6) Find the exact solutions of
ex+3e−x=4e^x+3e^{-x}=4


7) Find algebraically the exact solutions of
ln⁡(4−2x)+ln⁡(9−3x)=2ln⁡(x+1)\ln (4-2x)+\ln (9-3x)=2\ln (x+1)
where
−1<x<2-1<x<2


8) A curve has the equation
y=3ln⁡(2x−a)y=3\ln (2x-a)
, where
aa
is a positive constant. Sketch the curve and determine the
xx
-coordinate of
PP
in terms of
aa


9) Find the exact solution, in its simplest form, to the equation
3xe4x=e73^xe^{4x}=e^7


10) The point
PP
lies on the curve
y=4e2x+1y=4e^{2x+1}
. With
PP
having a
yy
-coordinate of 8, find, in terms of
ln⁡2\ln 2
, the
xx
-coordinate of
PP
Well done for completing this chapter! You can either have a go at the ECQs below, or head on to the next chapter.
1.ECQ End of Chapter Questions
ECQ
View the full catalogue of ECQs
This page was created by 9662e103-129a, and was last updated ~ May 2024