Continuous practice using OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Ex 6(D) can lead to a stronger grasp of mathematical concepts.

## S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Ex 6(D)

Solve the following equations :

Question 1.
2x+1 = 4x-3
Solution:
2x+1 = 4x-3 ⇒ 2x +1 = (22)x-3
⇒ 2x+1 = 22x-6
Comparing both sides, we get
x + 1 = 2x – 6 ⇒ 2x – x = 1 + 6
⇒ x = 7
∴ x = 7

Question 2.
$$x^{\frac{-3}{4}}=\frac{1}{8}$$.
Solution:
$$x^{\frac{-3}{4}}=\frac{1}{8}=\frac{1}{2^3}=2^{-3}$$
⇒ $$x^{\frac{1}{4} \times(-3)}=2^{-3}$$
⇒ $$\left(x^{\frac{1}{4}}\right)^{-3}=(2)^{-3}$$
Comparing,
⇒ x$$\frac { 1 }{ 4 }$$ = 2 ⇒ x = 24
⇒ x = 16
∴x = 16

Question 3.
(x – 1)$$\frac { 2 }{ 3 }$$ = 25
Solution:
(x – 1)$$\frac { 2 }{ 3 }$$ = 25 = 5²
(x – 1)$$\frac { 2 }{ 3 }$$ = (5³)$$\frac { 2 }{ 3 }$$
∴ Comparing, we get
x – 1 = 5³ ⇒ x – 1 = 125
⇒ x = 125 + 1 = 126
∴ x = 126

Question 4.
25x+3 = 8x + 3
Solution:
25x+3 = 8x + 3 = (2³)x+3
⇒ 25x+3 = 83x + 9
Comparing, we get
5x + 3 = 3x + 9
⇒ 5x – 3x = 9 – 3
⇒ 2x = 6 ⇒ x = $$\frac { 6 }{ 2 }$$ = 3
∴ x = 3

Question 5.
$$\left(\sqrt{\frac{5}{7}}\right)^{x-1}=\left(\frac{125}{343}\right)^{-1}$$
Solution:
$$\left(\sqrt{\frac{5}{7}}\right)^{x-1}=\left(\frac{125}{343}\right)^{-1}$$
⇒ $$\left(\frac{5}{7}\right)^{\frac{x-1}{2}}=\left(\frac{5^3}{7^3}\right)^{-1}=\left(\frac{5}{7}\right)^{-3}$$
Comparing both sides,
$$\frac { x-1 }{ 2 }$$ = – 3 ⇒ x – 1 = – 6
⇒ x = – 6 + 1 ⇒ x = – 5
∴ x = – 5

Question 6.
113-4x = $$\left(\sqrt{\frac{1}{121}}\right)^{-2}$$
Solution:
113-4x = $$\left(\sqrt{\frac{1}{121}}\right)^{-2}=\left(\sqrt{\frac{1}{(11)^2}}\right)^{-2}$$
⇒ 113-4x = 11$$\frac{(-2) \times(-2)}{2} \Rightarrow 11^{3-4 x}=11^2$$
Comparing, we get.
3 – 4x = 2 ⇒ – 4x = 2 – 3 = – 1
⇒ – 4x = – 1 ⇒ x = + $$\frac { 1 }{ 4 }$$
∴ x = $$\frac { 1 }{ 4 }$$

Question 7.
Solve for x, $$(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$$.
Solution:
$$(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$$
⇒ $$\left(\sqrt[3]{(2)^2}\right)^{2 x+\frac{1}{2}}=\frac{1}{2^5}$$
⇒ $$\left(2^{\frac{2}{3}}\right)^{2 x+\frac{1}{2}}$$ = 2-5
⇒ $$2^{\frac{2}{3} 2 x+\frac{1}{2}}$$ = 2-5
Comparing, we get
$$\frac{2}{3}\left(2 x+\frac{1}{2}\right)$$ = – 5
$$\frac{4}{3} x+\frac{1}{3}$$ = – 5
⇒ 4x + 1 = – 15
⇒ 4x = – 15 – 1 = – 16
⇒ x = $$\frac { -16 }{ 4 }$$ = – 4
∴ x = – 4

Question 8.
Find the value of .Y if $$\sqrt{\frac{p}{q}}=\left(\frac{q}{p}\right)^{1-2 x}$$
Solution:

Question 9.
Solve for x, 2³ (5° + 32x) = 8$$\frac { 8 }{ 27 }$$.
Solution:

Question 10.
Solve for x, $$\sqrt{\left(8^0+\frac{2}{3}\right)}$$ = (0.6)2-3x.
Solution:

Solve the following equations for x

Question 11.
32x+ 4 + 1 = 2.3x+ 2.
Solution:
32x+ 4 + 1 = 2.3x+ 2
⇒ 32x.34 + 1 = 2.33x.3²
⇒ 81.32x + 1 = 18.3x
⇒ 81.32x – 18.3x + 1 = 0
⇒ Let 3x = a, then 32x = a²
∴ 81a² – 18a + 1 = 0
⇒ (9a)² – 2 x 9a + (1)² = 0
⇒ (9a – 1)² = 0
⇒ 9a – 1 = 0 ⇒ 9a = 1
⇒ 9.3x = 1 ⇒ 3x = $$\frac { 1 }{ 9 }$$ = 3-2
Comparing both sides,
∴ x = – 2

Question 12.
52x+1 = 6.5x – 1.
Solution:
52x+1 = 6.5x – 1.
⇒ 52x.51 – 6.5x + 1 = 0
⇒ 5.52x – 6.5x + 1 = 0
Let 5x = a, then 52x = a²
∴ 5a² – 6a + 1 = 0
⇒ 5a² – 5a – a + 1 = 0
⇒ 5a (a – 1) – 1 (a – 1) = 0
⇒ (a – 1) (5a – 1) = 0
Either a – 1 = 0, then a = 1
or 5a – 1 = 0 then 5a = 1 ⇒ a = $$\frac { 1 }{ 5 }$$
(i) If a = 1, then 5x = 1 = 5° (∵ 5° = 1)
∴ x = 0
(ii) If a = $$\frac { 1 }{ 5 }$$ then 5x = $$\frac { 1 }{ 5 }$$ = 5-1
∴ x = – 1
Hence x = 0 or x = – 1

Question 13.
22x – 2x+3 = – 24
Solution:
22x – 2x+3 = – 24
⇒ 22x – 2x.23 + 24 = 0
⇒ 22x – 8.2x + 16 = 0
Let 22x = a, then 22x = a²
a² – 8a + 16 = 0
⇒ (a)² – 2 x a x 4 + (4)² = 0
⇒ (a – 4)² = 0
⇒ a – 4 = 0 ⇒ a = 4
∴ 2x = 4 = 2²
Comparing we get, x = 2

Solve for x and y :

Question 14.
9x = 3y-2, 81y = 3² x (27)x
Solution:
9x = 3y-2, 81y = 3² x (27)x
[(3)²]x = (3)y-2 ⇒ 32x = 3y-2
Comparing,
2x = y – 2
⇒ y = 2x + 2 … (i)
81y = 3² x (27)x ⇒ (34)y = 3² x (3³)x
⇒ (34)y = 3² x 33x ⇒ 34y = 33x+2
∴ 4y = 3x + 2 … (ii)
From (i)
4 (2x + 2) = 3x + 2
8x + 8 = 3x + 2 ⇒ 8x – 3x = 2 – 8
⇒ 5x = – 6 ⇒ x = $$\frac { -6 }{ 5 }$$
∴ y = 2x + 2 = 2 x ($$\frac { -6 }{ 5 }$$) + 2
= $$\frac { -12 }{ 5 }$$ + 2 = $$\frac{-12+10}{5}=\frac{-2}{5}$$
Hence x = $$\frac { -6 }{ 5 }$$ and y = $$\frac { -2 }{ 5 }$$

Question 15.
$$2^{1-\frac{x}{2}}=4^y,\left(7^{1+x}\right) \times(49)^{-2 y}$$ = 1
Solution:
$$2^{1-\frac{x}{2}}=4^y \Rightarrow 2^{1-\frac{x}{2}}=2^{2 y}$$
∴ 1 – $$\frac { x }{ 2 }$$ = 2y ⇒ 2 – x = 4y
⇒ 4y + x = 2 ⇒ x = 2 – 4y … (i)
and 71 + x x (49)-2y = 1
71+x x (7²)-2y = 7° (∵ 7° = 1)
⇒ 71 + x.7-4y = 7° ⇒ 71 + x-4y = 7°
∴ 1 + x – 4y = 0
⇒ 1 + (2 – 4y) – 4y = 0
From (i) x = 2 – 4y
⇒ 1 + 2 – 4y – 4y = 0 ⇒ 3 – 8y = 0
⇒ 8y = 3 ⇒ y = $$\frac { 3 }{ 8 }$$
∴ x = 2 – 4y = 2 – 4 x $$\frac { 3 }{ 8 }$$ = 2 – $$\frac { 3 }{ 2 }$$
= $$\frac { 4-3 }{ 2 }$$ = $$\frac { 1 }{ 2 }$$
Hence x = $$\frac { 1 }{ 2 }$$, y = $$\frac { 3 }{ 8 }$$

Question 16.
2x = 16 x 2y, (27)x = 9 x 32y
Solution:
2x = 16 x 2y ⇒ 2x = 24 x 2y
⇒ 2x = 24+y
Comparing, we get
∴ x = 4 + y … (i)
and (27)x = 9 x 32y ⇒ (3³)x = 3² x 32y
⇒ 33x = 32y+2
∴ 3x = 2y + 2
⇒ 3 (4 + y) = 2y + 2 [From (i)]
12 + 3y = 2v + 2 ⇒ 3y – 2y = 2 – 12
⇒ y = – 10
∴ x = 4 + 7 = 4 – 10 = – 6
Hence x = – 6, y = – 10