Selina Concise Class 9 Maths Chapter 7 Indices Exercise 7B Solutions
EXERCISE – 7B
(i) 22x+1 = 8
Solution:
22x+1 = 8
=> 22x+1 = 23
=> 2x + 1 = 3
=> 2x = 3 – 1
=> 2x = 2
=> x = 2/2
=> x = 1
(ii) 25x-1 = 4×23x+1
Solution:
25x-1 = 4 × 23x+1
2(5x-1) = 22 × 23x+1
25x-1 = 22+3x+1
25x-1 = 23+3x
=> 5x – 1 = 3 + 3x
=> 5x – 3x = 3 + 1
=> 2x = 4
=> x = 4/2
= x = 2
(iii) 34x+1 = (27)x+1
Solution:
34x+1 = (27)x+1
=> 34x+1 = (33)x+1
=> 34x+1 = 33(x+1)
=> 34x+1 = 33x+3
=> 4x + 1 = 3x + 3
=> 4x – 3x = 3 – 1
=> x = 2
(iv) (49)x+4 = 72 × (343)x+1
Solution:
(49)x+4 = 72 × (343)x+1
=> (72)x+4 = 72 × (73)x+1
=> 72(x+4) = 72 × 73 (x + 1)
=> 72x+8 = 72 × 73x+1
=> 72x+8 = 72+3x+1
=> 2x + 8 = 3x + 3
=> 2x – 3x = 3 – 8
=> -x = -5
=> x = 5
(Q2) Find x, if:
(i) 42x = 1/32
Solution:
42x = 1/32
=> (22)2x = 1/25
=> 24x = 2-5
=> 4x = -5
=> x = -5
(ii) √(2x+3) = 16
Solution:
√(2x+3) = 16
=> (2x+3)1/2 = 24
=> 21/2 (x + 3) = 24
=> 1/2 (x + 3) = 4
=> x + 3 = 2 × 4
=> x + 3 = 8
=> x = 8 – 3
=> x = 5
(iii) [√(3/5)]x+1 = 125/27
Solution:
[√(3/5)]x+1 = 125/27
=> ((3/5)1/2)x+1 = 53/33
=> (3/5)1/2(x+1) = (5/3)3
=> (3/5)1/2(x+1) = (3/5)-3
=> 1/2 (x + 1) = -3
=> x + 1 = 2 × (-3)
=> x + 1 = -6
=> x = -6-1
=> x = -7
(iv) [3√(2/3)]x-1 = 27/8
Solution:
[3√(2/3)]x-1 = 27/8
=> ((2/3)1/3)x-1 = 27/8
=> (2/3)1/3(x-1) = 33/23
=> (2/3)1/3(x-1) = (3/2)3
=> (2/3)1/3(x-1) = (2/3)-3
=> 1/3 (x – 1) = -3
=> x – 1 = (-3) × (3)
=> x – 1 = -9
=> x = -9+1
=> x = -8
(Q3) Solve:
(i) 4x-2 – 2x+1 = 0
Solution:
4x-2 – 2x+1 = 0
=> 4x-2 = 2x+1
=> (22)x-2 = 2x+1
=> 22(x-2) = 2x+1
=> 22x – 4 = 2x + 1
=> 2x – 4 = x + 1
=> 2x – x = 1 + 4
=> x = 5
(ii) 3x2 : 3x = 9:1
Solution:
3x2 : 3x = 9:1
(Q4) Solve:
(i) 8 × 22x + 4 × 2x+1 = 1 + 2x
Solution:
8 × 22x + 4 × 2x+1 = 1 + 2x
Let 2x = m
=> 8m2 + 8m = 1 + m
=> 8m2 + 8m – m – 1 = 0
=> 8m2 + 7m – 1 = 0
=> 8m2 + (8 – 1) m – 1 = 0
=> 8m2 + 8m – m – 1 = 0
=> 8m (m + 1) – 1 (m + 1) = 0
=> (m + 1) (8m – 1) = 0
=> m + 1 = 0
=> m = -1
= 2x = -1
Or, 8m – 1 = 0
Or, 8m = 1
m = 1/8
2x =1/8
2x = 1/23
2x = 2-3
x = -3
(ii) 22x + 2x+2 – 4×23 = 0
=> (2x)2 + 2x.22 – 4×8 = 0
Let 2x = m
=> m2 + 4m – 32 = 0
=> m2 + 8m – 4m – 32 = 0
=> m (m + 8) – 4 (m + 8) = 0
=> (m + 8) (m – 4) = 0
=> m + 8 = 0 or m – 4 = 0
=> m = -8 or m = 4
∴ 2x = -8 or 2x = 4
2x = 22
x = 2
(iii) (√3)x-3 = (4√3)x+1
Solution:
(√3)x-3 = (4√3)x+1
=> (3)1/2(x-3) = (3)1/4(x+1)
=> 1/2 (x – 3) = 1/4 (x + 1)
=> 4 (x – 3) = 2 (x + 1)
=> 4x – 12 = 2x + 2
=> 4x – 2x = 2 + 12
=> 2x = 14
x = 14/2
x = 7
(Q5) Find the values of m and n if:
42m = (3√16)-6/n = (√8)2
Solution:
Case I:
42m = (3√16)-6/n = (√8)2
42m = (√8)2
=> 22(2m) = (8)1/2×2
=> 24m = 23
=> 4m = 3
=> m = 3/4
Case II:
(3√16)-6/n = (√8)2
=> (16)1/3×-6/n = 81/2×2
=> (24)-2/n = 8
=> 24×-2/n = 23
=> -8/-n = 3
=> -8/3 = n
=> n = -8/13
(Q6) Solve x and y if:
(√32)x ÷ 2y+1 = 1 and 8y – 164-x/2 = 0
Solution:
(√32)x ÷ 2y+1 = 1
(√32)x/2(y+1) = 1
=> (32)1/2 x = 2y+1
=> (25)x/2 = 2y+1
=> (2)5×x/2 = (2)y+1
=> 5x/2 = y + 1 (∵ base same)
=> 5x = 2 (y + 1)
=> 5x = 2y + 2
=> 5x – 2y = 2 —– (i)
And 8y – 16(4-x/2) = 0
=> 8y = 16(4-x/2)
=> (23)y = (2)4(4-x/2)
=> (2)3y = (2)4(4-x/2)
=> (2)(3y) = (2)16-2x
=> 3y = 16 – 2x
=> 2x + 3y = 16 —– (ii)
Now, we have to solve equation (i) and (ii), simultaneously.
Multiply equation (i) by 2, we get,
2 (5x) – 2 × 2y = 2 × 2
10x – 4y = 4 —— (iii)
Now, multiply equation (ii) by 5, we get,
5 × 2x + 5 × 3y = 5 × 16
10x + 15y = 80 —— (iv)
∴ Subtracting equations (iii) – (iv), we get,
10x – 4y = 4
10x + 15y = 80
(-) (-) (-)
____________________
19y = 76
Y = 76/19
Y = 4
Put y = 4 in equation (i), we get,
5x – 2y = 2
5x – 2 (4) = 2
5x – 8 = 2
5x = 2 + 8
5x = 10
x = 10/5
x = 2
(Q7) Prove that,
(i) (xa/xb)a+b-c, (xb/xc)b+c-a.(xc/xa)c+a-b = 1
(ii)
(Q9) If ax = by = c and b2 = ac, prove that: y = 2az/(x + z)
Solution:
Proof:
ax = by = cz = k
ax = k
by = k
cz = k
ax = k
a = (k)1/x
by = k
b = (k)1/y
cz = k
c = (k)1/2
Now, b2 = ac
=> (k1/y)2 = (k1/x) . (k1/2)
=> (k1/y)2 = (k)1/x + 1/z
=> (k)2/y = (k)1/x + 1/z
=> 2/y = 1/x + 1/z
=> 2/y = z+x/xz
=> 2xz/z+x = y (Hence Proved)
(Q10) If 5-p = 4-q = 20r , show that 1/p + 1/q + 1/r = 0
Solution:
Proof: let 5-p = 4-q = 20r = k
5-p = k, 4-q = k, 20r = k
5 = (k)1/-p —— (i)
4 = (k)1/-q —– (ii)
20 = (k)1/r —— (iii)
Taking equation (iii),
20 = (k)1/r
=> 5×4 = (k)1/r
Now using equation (i) and (ii),
= (k)-1/p × (k)-1/q = (k)1/r
= (k)-1/p – 1/q = (k)1/r
= -1/p – 1/q = 1/r
= – (1/p + 1/q + 1/r) = 0
= 1/p + 1/q + 1/r = 0 (Hence Proved)
(Q11) If m ≠ n and (m + n)-1 (m-1 + n-1) = mx ny, show that: x + y + z = 0
Solution:
Proof:
(m + n)-1 (m-1 + n-1) = mxny
=> 1/m+n × (1/m + 1/n) = mxny
=> 1/(m + n) × (n + m/mn) = mxny
=> 1/mn = mxny
= m-1 n-1 = mx ny
=> x = -1 y = -1
Now, x + y + 2 = 0
LHS = x + y + 2
=> -1-1+2
=> -2+2
=> 0
∴ LHS = RHS (Hence proved)
(Q12) If 5x+1 = 25x-2, find the value of 3x-3 × 23-x
Solution:
Proof: 5x+1 = 25x-2
5x+1 = (52)x-2
5x+1 = 52(x-2)
X + 1 = 2x – 4
x + 1 = 2x – 4
x – 2x = -1 – 4
x – 2x = -5
-x = -5
X = 5
Now, 3x-3 × 23-x
= 35-3 × 23-5
= 32 × 2-2
= 9 × 1/22
= 9/4
(Q13) If 4x+3 = 112 + 8 × 4x, find the value of (18x)3x
Solution:
Proof:
4x+3 = 112 + 8 × 4x
4x. 43 – 8 × 4x = 112
4x . 64 – 8.4x = 112
4x × 8 (8 – 1) = 112
4x (7) = 112/8
4x . 7 = 14
4x = 14/7
4x = 2
(22)x = 2
22x = 21
2x = 1
X = 1/2
Now, we have to find,
(18x) 3x = (18 × 1/2)3×1/2
= (9)3/2
= (32)3/2
= 32×3/2
= (3)3
= 27
(Q14) Solve for x:
(i) 4x-1 × (0.5)3-2x = (1/8)-x
Solution:
(i) 4x-1 × (0.5)3-2x = (1/8)-x
(22)x-1 × (5/10)3-2x = (1/23)-x
22(x-1) × (1/2)3-2x = (2-3)-x
22x-2 × (2-1)3-2x = (2)3x
(2)2x-2 × (2)-3+2x = (2)3x
2x – 2 – 3 + 2x = 3x
4x – 5 = 3x
4x – 3x = 5
∴ x = 5
(ii)
(iii) (81)3/4 – (1/32)-2/5 + x (1/2)-1 × 20 = 27
Solution:
=> (81)3/4 – (1/32)-2/5 + x (1/2)-1 × 20 = 27
=> (34)3/4 – (1/25)-2/5 + x (2-1)-1 × 20 = 27
=> (3)4×3/4 – (2-5)-2/5 + x (2)(-1) ×(-1) × 1 = 27
=> (3)3 – (2)(-5) ×-2/(5) + x (2) × 1 = 27 (a0 = 1)
=> 27 – (2)2 + 2x = 27
=> 27 – 4 + 2x = 27
=> 23 + 2x = 27
=> 2x = 27 – 23
=> 2x = 4
=> x = 4/2
=> x = 2
(iv) 23x+3 = 23x+1 + 48
Solution:
23x+3 = 23x+1 + 48
=> 23x+3 – 23x+1 = 48
=> 23x × 23 – 23x × 21 = 48
=> 23x (23 – 2) = 48
=> 23x (8 – 2) = 48
=> 23x (6) = 48
=> 23x = 48/6
=> 23x = 8
=> 23x = 23
=> 3x = 3
=> x = 3/3
=> x = 1
(v) 3 (2x + 1) – 2x+2 + 5 = 0
Solution:
3 (2x + 1) – 2x+2 + 5 = 0
3 × 2x + 3 – 2x . 22 + 5 = 0
=> 2x × 3 – 2x . 22 + 8 = 0
=> 2x (3 – 4) + 8 = 0
=> 2x (-1) = -8
=> 2x = -8/-1
=> 2x = 8
=> 2x = 23
=> x = 3
(vi) 9x+2 = 720 + 9x
Solution:
9x+2 = 720 + 9x
=> 9x . 92 – 9x = 720
=> 9x (92 – 1) = 720
=> 9x (81 – 1) = 720
=> 9x × (80) = 720
=> 9x = 720/80
=> 9x = 91
=> x = 1
Here is your solution of Selina Concise Class 9 Maths Chapter 7 Indices Exercise 7B
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