C2.6 Exponentials and logarithms

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AS-Level Maths:
Core 2
for Edexcel
C2.6 Exponentials
and logarithms
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Exponential functions
Contents
Exponential functions
Logarithms
The laws of logarithms
Solving equations using logarithms
Examination-style questions
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Exponential functions
So far in this course we have looked at many functions
involving terms in xn.
In an exponential function, however, the variable is in the
index. For example:
x
y=2
x
y=5
y = 0.1
x
–x
y=3
x+1
y=7
The general form of an exponential function to the base a is:
y = ax where a > 0 and a ≠1.
You have probably heard of exponential increase and decrease
or exponential growth and decay.
A quantity that changes exponentially either increases or
decreases more and more rapidly as time goes on.
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Graphs of exponential functions
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Exponential functions
When a > 1 the graph of y = ax
has the following shape:
When 0 < a < 1 the graph of
y = ax has the following shape:
y
y
1
1
(1, a)
x
(1, a)
x
In both cases the graph passes through (0, 1) and (1, a).
This is because:
a0 = 1
and
a1 = a
for all a > 0.
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Logarithms
Contents
Exponential functions
Logarithms
The laws of logarithms
Solving equations using logarithms
Examination-style questions
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Logarithms
Find p if p3 = 343.
We can solve this equation by finding the cube root of 343:
p = 3 343
p =7
Now, consider the following equation:
Find q if 3q = 343.
We need to find the power of 3 that gives 343.
One way to tackle this is by trial and improvement.
Use the xy key on your calculator to find q to 2 decimal places.
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Logarithms
To avoid using trial and improvement we need to define the
power y to which a given base a must be raised to equal a
given number x.
This is defined as:
y = loga x
“y is equal to the logarithm, to the base a, of x”
The expressions y = loga x and
ay = x
are interchangeable.
This can be written using the implication sign :
y = loga x  ay = x
For example, 25 = 32 can be written in logarithmic form as:
log2 32 = 5
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Logarithms
Taking a log and raising to a power are inverse operations.
y = loga x  ay = x
We have that:
aloga x = x
So:
y = loga ay
Also:
For example:
log7 2
7
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=2
and
log3 36 = 6
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Contents
The laws of logarithms
Exponential functions
Logarithms
The laws of logarithms
Solving equations using logarithms
Examination-style questions
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Some important results
When studying indices we found the following important
results:
a1 = a
This can be written in logarithmic form as:
loga a = 1
a0 = 1
This can be written in logarithmic form as:
loga 1 = 0
It is important to remember these results when manipulating
logarithms.
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The laws of logarithms
The laws of logarithms follow from the laws of indices:
The multiplication law
Let:
m = loga x
and
n = loga y
So:
x = am
and
y = an

xy = am × an
Using the multiplication law for indices:
xy = am + n
Writing this in log form gives:
m + n = loga xy
But m = loga x and n = loga y so:
loga x + loga y = loga (xy)
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The laws of logarithms
The division law
Let:
m = loga x
So:
x = am
and
n = loga y
and y = an
x am
= n

y a
Using the division law for indices:
x
= a mn
y
Writing this in log form gives:
x
m  n  loga
y
But m = loga x and n = loga y so:
x
loga x  loga y  loga
y
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The laws of logarithms
The power law
Let:
m = loga x
So:
x = am

xn =(am)n
Using the power law for indices:
xn =amn
Writing this in log form gives:
mn = loga xn
But m = loga x so:
n loga x = loga xn
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The laws of logarithms
These three laws can be used to combine several logarithms
written to the same base. For example:
Express 2loga 3 + loga 2 – 2loga 6 as a single logarithm.
2loga 3 + loga 2  2loga 6 = loga 32 + loga 2  loga 62
= loga 9 + loga 2  loga 36
 9×2 
= loga 

36


= loga
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1
2
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The laws of logarithms
The laws of logarithms can also be used to break down a
single logarithm. For example:
2
Express log10
1
2
ab
in terms of log10 a, log10 b and log10 c.
4
c
2
1
2
1
ab
2 2
log10 4 = log10 a b  log10 c 4
c
1
2
2
= log10 a + log10 b  log10 c4
= 2log10 a + 21 log10 b  4log10 c
Logarithms to the base 10 are usually written as log or lg.
We can therefore write this expression as:
2log a + 21 log b  4log c
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Logarithms to the base 10 and to the base e
Although the base of a logarithm can be any positive number,
there are only two bases that are commonly used.
These are:
Logarithms to the base 10
Logarithms to the base e
Logarithms to the base 10 are useful because our number
system is based on powers of 10.
They can be found by using the log key on a calculator.
Logarithms to the base e are called Napierian or natural
logarithms and have many applications in maths and science.
They can be found by using the
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ln
key on a calculator.
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Changing the base of a logarithm
Suppose we wish to calculate the value of log5 8.
We can’t calculate this directly using a calculator because it
only find logs to the base 10 or the base e.
We can change the base of the logarithm as follows:
Let x = log5 8
5x = 8
So:
Taking the log to the base 10 of both sides:
log 5x = log 8
So:
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x log 5 = log 8
log 8
x=
log 5
log 8
log5 8 =
= 1.29 (to 3 s.f.)
log 5
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Changing the base of a logarithm
If we had used log to the base e instead we would have had:
ln 8
log5 8 =
= 1.29 (to 3 s.f.)
ln 5
In general, to find loga b:
Let x = loga b, so we can write ax = b
Taking the log to the base c of both sides gives:
logc ax = logc b
xlogc a = logc b
logc b
x=
logc a
So:
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logc b
loga b =
logc a
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Contents
Solving equations using logarithms
Exponential functions
Logarithms
The laws of logarithms
Solving equations using logarithms
Examination-style questions
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Solving equations involving logarithms
We can use the laws of logarithms to solve equations.
For example:
Solve log5 x + 2 = log5 10.
To solve this equation we have to write the constant value 2 in
logarithmic form:
2 = 2 log5 5
because log5 5 = 1
= log5 52
= log5 25
The equation can now be written as:
log5 x + log5 25 = log5 10
log5 25x = log5 10
25x = 10
x = 0.4
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Solving equations of the form ax = b
We can use logarithms to solve equations of the form ax = b.
For example:
Find x to 3 significant figures if 52x = 30.
We can solve this by taking logs of both sides:
log 52x = log 30
Using a calculator:
2x log 5 = log 30
log 30
2x =
log 5
log 30
x=
2log 5
x = 1.06 (to 3 s.f.)
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Solving equations of the form ax = b
Find x to 3 significant figures if 43x+1 = 7x+2.
Taking logs of both sides:
log 43 x 1 = log 7 x  2
(3 x +1)log 4 = ( x + 2)log 7
3 x log 4 +log 4 = x log 7 + 2log 7
x(3log 4  log 7) = 2log 7  log 4
2log 7  log 4
x=
3log 4  log 7
x = 1.13 (to 3 s.f.)
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Solving equations of the form ax = b
Solve 32x –5(3x) + 4 = 0 to 3 significant figures.
If we let y = 3x we can write the equation as:
y2  5 y + 4 = 0
( y  1)( y  4) = 0
y =1 or y = 4
So:
3x =1 or 3x = 4
If 3x = 1 then x = 0.
Now, solving 3x = 4 by taking logs of both sides:
log 3 x = log 4
xlog 3 = log 4
log 4
x=
log 3
x = 1.26 (to 3 s.f.)
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Contents
Examination-style questions
Exponential functions
Logarithms
The laws of logarithms
Solving equations using logarithms
Examination-style questions
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Examination-style question
Julia starts a new job on a salary of £15 000 per annum. She
is promised that her salary will increase by 4.5% at the end of
each year. If she stays in the same job how long will it be
before she earns more than double her starting salary?
15 000 × 1.045n = 30 000
1.045n = 2
log 1.045n = log 2
n log 1.045 = log 2
log 2
n=
 15.7
log 1.045
Julia’s starting salary will have doubled after 16 years.
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