1) The document provides instructions for solving various mathematical equations and expressions involving indices. 2) Examples include solving equations for specific values of x, expressing one term in terms of others, and solving simultaneous equations. 3) Key steps shown include isolating variables, applying inverse operations, and setting expressions equal to find solutions.

1. Cikgu Hanini Suhaila Hamsan Indices puterizamrud@gmail.com/ April2018
INDICES
𝑎 𝑚
𝑎 𝑛
= 𝑎 𝑚−𝑛
(𝑎 𝑚) 𝑛 = 𝑎 𝑚𝑥𝑛 = 𝑎 𝑚𝑛
𝑎 𝑥x 𝑏 𝑥= (ab) 𝑥
𝑎0= 1
𝑎1= a
𝑎−1 =
1
𝑎1
𝑎 𝑚x 𝑎 𝑚= 𝑎 𝑚+𝑛
1 Solve the equation 2 Solve the equation
9 𝑥= 9(3 𝑥) -18 3 𝑥+2 - 3 𝑥= 24
32𝑥 = 32(3 𝑥)- 18 3 𝑥 . 32 − 3 𝑥= 24
(3 𝑥)2 - 32(3 𝑥)- 18 = 0 Substitute y=3 𝑥 3 𝑥 (32 − 1 )= 24
𝑦2- 9y – 18 = 0 3 𝑥 (8)= 24
(y- 3 ) ( y - 6 ) =0 1x 18 , 2X9 , 6 x 3 3 𝑥 = 24/8
Y= 3 or y = 6 3 𝑥 = 3
3 𝑥= 3 3 𝑥 = 6 X= 1
X= 1 lg 3 𝑥= lg6
X lg3= lg6 Or y= 3 𝑥 𝑦 ( 8 ) = 24 𝑦 = 3
X =
lg 6
lg 3
=
0.778
0.477
= 1.631
3 𝑥 = 3
X = 1
3 Solve the simultaneousequation
23𝑥
44𝑦
= 32 and 5 𝑥 . 25 𝑦−1 = 3 125
23𝑥
44𝑦
= 32
23𝑥 ÷ 22𝑦 = 32 3x - 2y = 5………….(1) Whenx = 3, 3 + 2y=7
23𝑥−2𝑦 = 25 + X+ 2y = 7………….(2) 2y=4
Y=2
3x-2y = 5………….(1) 4x = 12
X = 3
4 5 𝑥 . 25 𝑦−1 = 3 125
5 𝑥 . 52(𝑦−1) = 55
5 𝑥+2𝑦−2 = 55
X+ 2y -2 = 5
X+ 2y= 7………….(2)

2. Cikgu Hanini Suhaila Hamsan Indices puterizamrud@gmail.com/ April2018
5
5 𝑎 = 11 𝑏 = 55 𝑐
Expressc intermsof a
6
272𝑥−5 =
1
√9 𝑥+1
3 3(2𝑥−5) = 32(𝑥+1) ) −1/2
5 𝑎 = 11 𝑏 = 55 𝑐= k =3−(𝑥+1) )
5 𝑎 = 𝑘 11 𝑏 = 𝑘 55 𝑐= k 3 ( 2x-5) = - ( x + 1)
5= 𝑘1/𝑎 11 = 𝑘1/𝑏 55 = 𝑘1/𝑐 6x-15 = -x -1
6x+x = -1 +15
5 x 11= 55 7x= 14
𝑘1/𝑎x 𝑘1/𝑏= 𝑘1/𝑐
1
𝑎
+
1
𝑏
=
1
𝑐
𝑏+𝑎
𝑎𝑏
=
1
𝑐
𝑐
1
=
𝑎𝑏
𝑏 + 𝑎
X= 2
7 2 𝑥+3 + 2 𝑥 + 16(2 𝑥−1)
to the form k (2 𝑥)
= 2 𝑥.2 3 + 2 𝑥 + 16(2 𝑥) . 2−1
= 2 𝑥 ( 8 + 1 + 16(1/2) )
= 2 𝑥( 17)
8 Findthe value y
(6 𝑦 ) 2 .
1
216 𝑦
= 36
(6 𝑦 ) 2 .
1
6 3𝑦
= 62
(6 𝑦 ) 2.6 −3𝑦 = 62
2y -3y = 2
-y = 2
Y = -2
9 Solve 5 𝑥2
− 25 6−2𝑥 = 0
5 𝑥2
= 5 2(6−2𝑥)
𝑥2=2 (6-2x)
𝑥2 = 12 − 4𝑥
𝑥2 + 4𝑥 − 12 = 0
( x + 6 ) (x - 2 ) = 0
X= -6 , x=2