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Pointers

•

Since 27 and 9 are both powers of 3, both sides of the equation can be expressed in indexform as powers of 3.

• When the bases on both sides of the equation are the same, the indices on both sides of theequation must be the same.

•

The equation is then solved by equating the indices.

Chapter 5 : Indices and Logarithms

Cloned SPM Question (Paper 1)

Solve the equation 3 5

3

127

9

x

x

−

−

= .

Solution

3 5

3

127

9

x

x

−

−

=

3(3 5)

12(3 ) 2

13

3

x

x

−

−

=

(3 )

1

3 x−=

(3 )3 x− −

=

3(3 x − 5) = −(3 − x)

9 x − 15 = x − 3

8 x = 12

x =3

2

Equate the indices.

NEXUS VISTA SPM ADDITIONAL MATHEMATICS FORM 4 & 5

Page 67 (More Cloned SPM Questions) Supplementary Materials for

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Pointers

•

Since all the terms involving logarithms are to the same base, that is base 3, they can be

combined as a single term using the laws of logarithms.

• The equation is then changed into index form. Then make p the subject of the equation.

Pointers

•

Combine all terms involving logarithms into a single term using the laws of logarithms.

•

Then, change the equation from the logarithmic form to the index form, which can be easily

solved.


Given 3 3 3log 2 3 log log pq p q= + − , express p in terms of q.

Solution 3 3 3log 2 3 log log pq p q= + −

3

3 3 3log log log 2 pq p q− + =

2

3 3log 2

pq

p=

22

23

q

p=

3

q p =


Solve the equation ( )2 23 log 1 log x x+ − = .

Solution

( )2 23 log 1 log x x+ − =

( )2 23 log log 1 x x= − −

2log 31

x

x=

−

321

x

x=

−

8 8 x x= −

7 x = 8 x =

8

7



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Pointers

• Observe that the given logarithms are to the base 3 while the given expression is to the

base 9. So, change the logarithm to the base 9 to logarithms to the base 3 using the

formulaa

bb

c

ca

log

loglog = .

•

Next, apply the laws of logarithms on product and quotient.

• Since 27 and 9 are both powers of 3, their logarithms to the base 3 are 3 and 2 respectively.


Given 3log p m= and 3log q n= , express 9

27log

p

q

in terms of m and n.

Solution

3

93

27log

27log

log 9

p

q p

q

=

= 3 3 3

23

log 27 log log

log 3

p q+ −

=3

3log 3

2

m n+ −

= 12

(3 + m – n)



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Pointers

• Observe that all the numbers in this equation are powers of 2.

• Use the laws of indices to express: 2

3(2

n – 2) as 2

3 + n – 2

(24)n as 2

4n

• When both sides of the equation are numbers in index form with the same base, we can

equate the indices and solve the equation.


Given 8(2n – 2

) = 16n, find the value of n.

Solution

8(2n – 2

) = 16n

23(2

n – 2) = (2

4)n

23 + n – 2

= 24n

2n + 1

= 24n

n + 1 = 4n

3n = 1

n =3

1



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Pointers

• To solve any logarithmic equation, first change the bases of all the logarithms of the

equation to a common base.

•

Observe that the terms in the given logarithmic equation are to the bases 81 and 3.

•

Since 81 is a power of 3, change log 81 x

2

to the base 3 by using the formula a

b

b c

c

a log

log

log =

.

•

Note that log 3 81 = 4 because log 3 81 = log 3 34 = 4 log 3 3 and log 3 3 = 1.


Given log 81 x2 = log 3 5, find the value of x.

Solution

log 81 x2 = log3 5

81log

log

3

2

3 x = log3 5

4

log2 3 x = log 3 5

2

log3 x = log 3 5

log 3 x = log 3 5

2

Thus, x = 52

= 25



pg_0117_0067

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