The document RD Sharma Solutions Ex-2.2, (Part - 1), Exponents Of Real Numbers, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**1. Assuming that x,y,z are positive real numbers, simplify each of the following**

**2. Simplify**

**3. Prove that**

**4. Show that**

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

Therefore, LHS = RHS

Hence proved

(ii)

Left hand side (LHS) = Right hand side (RHS)

Considering LHS,

1

Therefore, LHS = RHS

Hence proved

**5. If 2 ^{x} = 3^{y} = 12^{z} , **

**6. If 2 ^{x} = 3^{y} = 6^{−z} , **

[by equating exponents]

* * **7. If a ^{x} = b^{y} = c^{z} and b^{2} = ac , then show that **

**8. If 3 ^{x }= 5^{y }= (75)^{z}, show that **

We have,

3x = 2 − x [On equating exponents]

3x + x = 2

4x = 2

x = 2/4

x = 12

Here the value of x is 1/2

**10. Find the values of x in each of the following**

We have

= 4x = 4 [On equatinge xponent]

x = 1

Hence the value of x is 1

(ii). (2^{3})^{4} = (2^{2})^{x}

We have

(2^{3})^{4} = (2^{2})^{x}

= 2^{3×4} = 2^{2×x}

12 = 2x

2x = 12 [On equating exponents]

x = 6

Hence the value of x is 6

We have

x = 3 [on equating exponents]

Hence the value of x is 3

(iv) 5^{x−2}×3^{2x−3 }= 135

We have,

⇒ x − 2 = 1,2x − 3 = 3 [On equating exponents]

⇒ x = 2 + 1,2x = 3 + 3

⇒ x = 3, 2x = 6 ⇒ x = 3

Hence the value of x is 3

(v). 2^{x−7}×5^{x−4 }= 1250

We have

2^{x−7}×5^{x−4} = 1250

⇒ 2^{x−7}×5^{x−4 }= 2×625

⇒ 2^{x−7×}5^{x−4 }= 2×5^{4}

⇒ x−7 = 1 ⇒ x = 8,x−4 = 4 ⇒ x = 8

Hence the value of x is 8

4x+1 = −15

4x = −15−1

4x = −16

x =−16/4

x = −4

Hence the value of x is 4

(vii). 5^{2x+3 }= 1

5^{2x+3 }= 1×5^{0}

2x+3 = 0 [By equatinge xponents]

2x = −3

x = −3/2

Hence the value of x is −3/2

[By equating exponents]

Hence the value of x is 4

Hence the value of x is 7

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