Logarithms in Mathematics, Slides of Mathematics

The concept of logarithms in mathematics. It defines the base and logarithm of an expression and provides examples of how to evaluate logarithms. It also lists the logarithm laws and identities that hold for positive numbers and real numbers. The document concludes by introducing the natural logarithm and its properties.

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The University of New South Wales
School of Mathematics and Statistics
Mathematics Drop–in Centre
LOGARITHMS
“Logarithm” is really just another word for “exponent” or
“power”. In an expression abwe call athe base and bthe loga-
rithm. So if
ab=c
then we say that bis the logarithm of cto the base a, written
b= logac.
In all of these equations, aand cmust be positive numbers; b
may be positive, negative or zero. In principle, logarithms can be
evaluated by rewriting them as powers.
Example. Evaluate log232.
Solution. Let x= log232. Then 2x= 32 and so by trial and
error x= 5.
In practice, evaluating logarithms usually requires a calculator.
For instance, if x= log234 then 2x= 34; the previous example
shows that xmust be a bit more than 5 and definitely less than
6, but it’s difficult to pin it down much more closely than this.
Logarithm laws. The following identities hold, where a, x, y are
positive numbers and pis any real number:
loga(xy) = logax+ logay
logax
y= logaxlogay()
loga(xp) = plogax .
In these equations it does not matter what base we use for the
logarithms, so long as it is the same throughout the equation.
So we shall often write simply log(xy) = log x+ log yand so on.
Combining the above properties we have, for example,
7 log(x2y5)9 log(xy4)
= 7(2 log x+ 5 log y)9(log x+ 4 log y)
= 5 log xlog y
= logx5
y.
Note that there is no useful simplification for log(x±y).
Further identities follow from the definition of the logarithm:
alogax=xand loga(ax) = x .
In this case the base of the logarithm does matter and cannot be
omitted. A sample simplification:
100log10 x=10log10 x2
=x2.
An especially important logarithm is the one with base e=
2·71828 ···. It is called the natural logarithm and written ln.
That is, ln xmeans the same as logex. All of the identities ()
remain true when logais replaced by ln. For example,
lne3
x= ln(e3)ln(x1/2) = 3 ln e1
2ln x= 3 1
2ln x .
Comment. We have ignored certain difficulties in defining loga-
rithms, and a more careful approach will be taken in MATH1131
lectures. Nevertheless, the properties given will remain true and
we can still manipulate expressions as in the previous examples.
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The University of New South Wales School of Mathematics and Statistics Mathematics Drop–in Centre

LOGARITHMS

“Logarithm” is really just another word for “exponent” or

“power”.

In an expression

a b we call

a

the base and

b

the loga-

rithm. So if

a b =

c

then we say that

b is the logarithm of

c to the base

a , written

b = log

a c .

In all of these equations,

a

and

c

must be positive numbers;

b

Exampleevaluated by rewriting them as powers. may be positive, negative or zero. In principle, logarithms can be

. Evaluate log

2 (^) 32.

Solution

Let

x

= log

2

Then 2

x

= 32 and so by trial and

error

x

= 5.

For instance, ifIn practice, evaluating logarithms usually requires a calculator.

x

= log

2 34 then 2

x

= 34; the previous example

shows that

x

must be a bit more than 5 and definitely less than

Logarithm laws6, but it’s difficult to pin it down much more closely than this.

. The following identities hold, where

a, x, y

are

positive numbers and

p

is any real number:

log

a ( xy

) = log

a x (^) + log

a y

log

a ( y x^ )

= log

a x (^) −

(^) log

a y

log

a ( x p ) =

p (^) log

a

x.

So we shall often write simply log(logarithms, so long as it is the same throughout the equation. In these equations it does not matter what base we use for the

xy

) = log

(^) x

  • log

(^) y

and so on.

Combining the above properties we have, for example,

7 log(

x 2 y 5 ) −

9 log(

xy

4 )

= 7(2 log

(^) x (^) + 5 log

(^) y ) (^) −

(^) 9(log

(^) x

  • 4 log

(^) y )

= 5 log

(^) x

log

(^) y

= log

x^ 5

y

)

.

Note that there is

no useful simplification

for log(

x ±

y ).

Further identities follow from the definition of the logarithm:

a log

a (^) x

=

x

and

log

a ( a x ) =

x.

In this case the base of the logarithm

does

matter and cannot be

omitted. A sample simplification:

log

10 (^) x

=

( 10

log

10 (^) x ) 2 = x 2.

An especially important logarithm is the one with base

e

=

(^). It is called the

natural logarithm

and written ln.

That is, ln

(^) x

means the same as log

e (^) x .

All of the identities (

remain true when log

a

is replaced by ln. For example,

ln

(

e 3

x )

= ln(

e 3 ) −

ln(

x 1 / 2 ) = 3 ln

(^) e (^) −

21 ln

(^) x

21 ln

(^) x.

Comment

. We have ignored certain difficulties in defining loga-

we can still manipulate expressions as in the previous examples.lectures. Nevertheless, the properties given will remain true andrithms, and a more careful approach will be taken in MATH

EXERCISES

Please try to complete the following exercises.

Remember that

you

cannot

expect to understand mathematics without doing lots

of practice!

Please do not look at the answers before trying the

please consult your tutor or the Mathematics Drop–in Centre.which you cannot find, or a question which you cannot even start,working carefully, find the mistake and fix it. If there is a mistakequestions. If you get a question wrong you should go through your

  1. Write the following statements as equations involving loga-

rithms: (a) 1331 = 11

3 ;

(b) 25

1 / 2 = 5;

(c)

y

x π ;^

(d)

p q = 7;

(e) 2

x

=

23 (^) ;

(f)

z − 1 · 23

  1. Write the following as equations involving powers:

(a) 7 = log

3 2187;

(b)

x

= log

10

2;

(c) log

5 (^) a

=

(d)

x

= ln 5;

(e) ln

(^) x

= 5.

  1. Evaluate, without using a calculator,

(a) log

2 128;

(b) log

125

(c) log

10

1

100

  1. Simplify if possible

(a) log

2 ( x^ 2

y 5 )

  • log

2 ( 8 y 7

x 9 ) ;

(b) 3

log

3 x − 2 log

3 y ;^

(c) log(

x 2

y 2 );

(d) 4 ln

s^ 5

t 6 )

  • ln

s^ 7

t 8 ) ;

(e) log

18

(

z 3 2 z );^

(f) (ln

(^) a )(ln

(^) b );

(g) ln 10 + ln 100 + ln 1000;

(h) ln

xy

− 3 ) −

ln

( x^ 6

y 7 ) .

ANSWERS

(a) log

11

(^) 1331 = 3;

(b) log

25

(^) 5 =

21 (^) ;

(c) log

x (^) y

=

π ;

(d) log

p 7 =

q ;

(e)

x

= log

2 23 (^) ;

(f) log

z 4 · 56 =

(a) 3

7 = 2187;

(b) 10

x = 2;

(c)

a = 5

− 3 ;

(d)

e x

= 5;

(e)

x

=

e 5 .

(a) 7;

(b)

31 (^) ;

(c)

(a) 3

(^) 7 log

2 (^) x

  • 2 log

2 (^) y , or 3 + log

2 ( y^ 2

x 7 ) ;

(b)

x y 2 (^) ;

(d) 27 ln(c) no simplification;

(^) s

32 ln

(^) t , or ln

s^ 27

t 32

(e)

z ;

(f) no simplification; the given expression can be written as

ln(

b ln (^) a ) or ln(

a ln (^) b ), but these are not really simpler;

(h) (g) 6 ln 10;

9 ln

(^) x (^) + 4 ln

(^) y , or ln

y^ 4

x 9 ) .