Indices Questions and Answers - Form 2 Topical Mathematics

1. Evaluate the value of x in
81(x+1) + 34x = 246
2. Solve for y in the equation:-
5(2y+1) = 4(5)y+1 − 15
3. Without logarithm tables or calculators, evaluate: (25)¾ x 0.92 x 22/55⁄2 x 33 in the form A/B where A and B are integers
4. Find the value of x given that :
2x=0.0625 (x is an integer)
5. Find the value of x which satisfies the equation 16x2 = 8(4x – 3)
6. Solve the equation;
9(x+1) + 3(2x+1)= 36
7. By letting P = 4-y in the equation:
4(-2y+1) − 3 x 4-y − 10 = 0
1. Write the above equation in terms of P
2. Hence find the possible values of y
8. Solve for x in the equation
1/16x = 1/32
9. In the expansion of [ax − 2/x2]6 the constant term is 4860. Find the value of a.

1. 34 x 34x + 34x = 246
34x (81 + 1 ) = 246
82/82 x 34x = 246/82
34x = 31    √
4x = 1
x = 1/4
2. 52y x51 = 4(5y+1) − 15
5y x 5y x 51 = 4 x 5y x 51 − 15
Let 5y = t
5t2 = 20t − 15
t2 = 20t − 15
t2 − 4t + 3 = 0
(t-1) (t-3) = 0
t = 1 or 3
5y = 1= 5o
Or 5y = 3y = log3
log5 = 0.6826
3. 25¾ = (25½)3/2  = 5
0.92 = (9/10)2 = 92/100
22 = 22
(√5)3 x 92 x 2/ (√5)5 x 102 x 33
3 x 4/(√5)2 x 102
3/(5 X 25) = 3/125
4. 2x = 0.0625 = 625/1000
2x = 1/16 = 2-4
x = -4
5. 16x2 = 8(4x-3)
2(4x2) = 2 3(4x -3)
= 4x2 = 12x -9
= 4x2 − 12x + 9 = 0
(2x−3)2 = 0
2x−3 = 0
x = 1.5
6.  9 (x+ 1) + 3(2x+1) = 36

3(2x+2) + 3(2x+ 1) = 36
32x(9+3) = 36
32x = 31
2x = 1
x = ½

1. 4p2 − 3P −10 =0
2. 4p2 − 8p + 5p =0
(4p +5) (p−2) = 0
p1 = -5/4, p =2
When y = -5/4,
4-y = -5/4
y = log4 (-5)/ 2
P = 2
4-y = 2
2-2y = 21
y = -1/2
7. 1/16x = 1/32
[1/24x ](x- ¼)= 1/25

2(-4xˆ2 + x) + x = 2-5
-4x2 + x + 5 = 0
4x2 − x − 5 = 0
4x2 −5x + 4x – 5 = 0
x(4x – 5) + 1(4x – 5) = 0
x = -1 or x = 5/4
8. 15 (ax)4 (-2/x2) = 4860
60a4 = 4860
a4 = 81
a = 3

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