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AS-Level Maths: Core 2for Edexcel

C2.6 Exponentials and logarithms

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Exponential functions

Logarithms

The laws of logarithms

Solving equations using logarithms

Examination-style questions




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:

The general form of an exponential function to the base a is:

y = ax where a > 0 and a ≠1.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.

y = 2x y = 5x y = 0.1x y = 3–x y = 7x+1


Graphs of exponential functions



In both cases the graph passes through (0, 1) and (1, a).This is because:

a0 = 1 and a1 = afor all a > 0.

When 0 < a < 1 the graph ofy = ax has the following shape:

y

x

1 1

When a > 1 the graph of y = ax has the following shape:

y

x

(1, a)(1, a)


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Logarithms




Logarithms


Logarithms

Find p if p3 = 343.

We can solve this equation by finding the cube root of 343:3= 343p

= 7p

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.


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”

This can be written using the implication sign :

y = loga x ay = x y = loga x ay = x

The expressions y = loga x ay = xand are interchangeable.

For example, 25 = 32 can be written in logarithmic form as:

log2 32 = 5


Logarithms

Taking a log and raising to a power are inverse operations.

We have that: y = loga x ay = x y = loga x ay = x

So: log =a xa x

Also: y = loga ayy = loga ay

For example:

7log 27 = 2 and 63log 3 = 6


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Logarithms






Some important results

When studying indices we found the following important results:

This can be written in logarithmic form as:

loga a = 1loga a = 1

a1 = a

This can be written in logarithmic form as:

loga 1 = 0loga 1 = 0

a0 = 1

It is important to remember these results when manipulating 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

loga x + loga y = loga (xy)loga x + loga y = loga (xy)

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:



The division lawLet: m = loga x and n = loga y

So: x = am and y = an

Using the division law for indices:


But m = loga x and n = loga y so:

=m

n

x a

y a

= m nxa

y

loga

xm n

y

log log loga a a

xx y

y



The power law

Let: m = loga x

So: x = am

Using the power law for indices:


But m = loga x so:

xn =(am)n

xn =amn

mn = loga xn

n loga x = loga xnn loga x = loga xn



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.

2 2log 3 + log 2 log 6a a a2log 3 + log 2 2log 6 =a a a

= log 9 + log 2 log 36a a a

9×2= log

36a

12= loga



Express log10 in terms of log10 a, log10 b and log10 c.122

4

a b

c122

10 4log =

a b

c

122 4

10 10log loga b c

a b c122 4

10 10 10= log + log log

a b c110 10 102= 2log + log 4log

Logarithms to the base 10 are usually written as log or lg.

We can therefore write this expression as:

a b c122log + log 4log

The laws of logarithms can also be used to break down a single logarithm. For example:


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 ln key on a calculator.


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

So: 5x = 8

Taking the log to the base 10 of both sides:

log 5x = log 8

x log 5 = log 8

xlog 8

=log 5

So: 5

log 8log 8 = =

log 51.29 (to 3 s.f.)


Changing the base of a logarithm

If we had used log to the base e instead we would have had:

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

log=

logc

c

bx

a

5

ln 8log 8 = =

ln 51.29 (to 3 s.f.)

loglog =

logc

ac

bb

aSo:


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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


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

2x log 5 = log 30

xlog 30

2 =log 5

xlog 30

=2log 5

Using a calculator:

x = 1.06 (to 3 s.f.)



Find x to 3 significant figures if 43x+1 = 7x+2.

Taking logs of both sides:x x 3 1 2log 4 = log 7

x x(3 +1)log 4 = ( + 2)log 7

x x3 log 4 + log 4 = log 7 + 2log 7

x (3log 4 log 7) = 2log 7 log 4

x

2log 7 log 4=

3log 4 log 7

=1.13 (to 3 s.f.)x



Solve 32x –5(3x) + 4 = 0 to 3 significant figures.

If we let y = 3x we can write the equation as:2 5 + 4 = 0y y

( 1)( 4) = 0y y =1 or = 4y y

So: 3 =1 or 3 = 4x x

If 3x = 1 then x = 0.Now, solving 3x = 4 by taking logs of both sides:

xlog 3 = log 4x log 3 = log 4

xlog 4

=log 3

=1.26 (to 3 s.f.)x


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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=

log 1.045n 15.7

Julia’s starting salary will have doubled after 16 years.

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