OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Ex 6(D)

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.
2^x+1 = 4^x-3
Solution:
2^x+1 = 4^x-3 ⇒ 2^{x +1} = (2²)^x-3
⇒ 2^x+1 = 2^2x-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.
2^5x+3 = 8^{x + 3}
Solution:
2^5x+3 = 8^{x + 3} = (2³)^x+3
⇒ 2^5x+3 = 8^{3x + 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.
11^3-4x = \(\left(\sqrt{\frac{1}{121}}\right)^{-2}\)
Solution:
11^3-4x = \(\left(\sqrt{\frac{1}{121}}\right)^{-2}=\left(\sqrt{\frac{1}{(11)^2}}\right)^{-2}\)
⇒ 11^3-4x = 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° + 3^2x) = 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.
3^{2x+ 4} + 1 = 2.3^{x+ 2}.
Solution:
3^{2x+ 4} + 1 = 2.3^{x+ 2}
⇒ 3^2x.3⁴ + 1 = 2.3^3x.3²
⇒ 81.3^2x + 1 = 18.3^x
⇒ 81.3^2x – 18.3^x + 1 = 0
⇒ Let 3^x = a, then 3^2x = a²
∴ 81a² – 18a + 1 = 0
⇒ (9a)² – 2 x 9a + (1)² = 0
⇒ (9a – 1)² = 0
⇒ 9a – 1 = 0 ⇒ 9a = 1
⇒ 9.3^x = 1 ⇒ 3^x = \(\frac { 1 }{ 9 }\) = 3^-2
Comparing both sides,
∴ x = – 2

Question 12.
5^2x+1 = 6.5^x – 1.
Solution:
5^2x+1 = 6.5^x – 1.
⇒ 5^2x.5¹ – 6.5^x + 1 = 0
⇒ 5.5^2x – 6.5^x + 1 = 0
Let 5^x = a, then 5^2x = 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 5^x = 1 = 5° (∵ 5° = 1)
∴ x = 0
(ii) If a = \(\frac { 1 }{ 5 }\) then 5^x = \(\frac { 1 }{ 5 }\) = 5^-1
∴ x = – 1
Hence x = 0 or x = – 1

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

Solve for x and y :

Question 14.
9^x = 3^y-2, 81^y = 3² x (27)^x
Solution:
9^x = 3^y-2, 81^y = 3² x (27)^x
[(3)²]x = (3)^y-2 ⇒ 3^2x = 3^y-2
Comparing,
2x = y – 2
⇒ y = 2x + 2 … (i)
81^y = 3² x (27)^x ⇒ (3⁴)^y = 3² x (3³)^x
⇒ (3⁴)^y = 3² x 3^3x ⇒ 3^4y = 3^3x+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 7^{1 + x} x (49)^-2y = 1
7^1+x x (7²)^-2y = 7° (∵ 7° = 1)
⇒ 7^{1 + x}.7^-4y = 7° ⇒ 7^{1 + 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.
2^x = 16 x 2^y, (27)^x = 9 x 3^2y
Solution:
2^x = 16 x 2^y ⇒ 2^x = 2⁴ x 2^y
⇒ 2^x = 2^4+y
Comparing, we get
∴ x = 4 + y … (i)
and (27)^x = 9 x 3^2y ⇒ (3³)^x = 3² x 3^2y
⇒ 3^3x = 3^2y+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