Exponents of Real Numbers- 1 RD Sharma Solutions | Mathematics (Maths) Class 9 PDF Download

RD Sharma Solutions Exercise 2.1 Exponents Of Real Numbers

(i) 3(a⁴b³)¹⁰ × 5(a²b²)³

(ii) (2x⁻²y³)³

(iii)

(iv)

(v)

(vi)

Proof: (i)

3(a⁴b³)1⁰×5(a²b²)³

=3×a⁴⁰×b³⁰×5×a⁶×b⁶

=15×a⁴⁰×a⁶×b³⁰×b⁶

=15×a⁴⁰⁺⁶×b³⁰⁺⁶              [a^m×aⁿ=a^m+n]

=15a⁴⁶b³⁶

(ii)

(2x⁻²y³)³

=2³×(x⁻²)³×(y³)³

=8×x⁻⁶×y⁹

=8x⁻⁶y⁹

(iii)

=3×10²⁺⁽⁻⁴⁾

=3×10⁻²

=

(iv)

=−2×a²×b⁵×a⁻²×b⁻²
=−2×a²⁺⁽⁻²)×b⁵⁺⁽⁻²⁾

=−2×a⁰×b³

=−2b³

(v)

(vi)

=a^(18n−54)×a^−(2n−4)

=a¹⁸ⁿ⁻⁵⁴×a⁻²ⁿ⁺⁴

=a^{18n−54−2n+4}

=a¹⁶ⁿ⁻⁵⁰

Q.2. If a=3 and b=−2, find the values of:

(i) a^a+b^b

(ii) a^b+b^a

(iii) (a+b)^ab

Proof: (i) a^a+b^b

Here, a=3 and b=−2.

Put the values in the expression a^a+b^b.

3³+(−2)⁻²

=27+1/(−2)²

=27+1/4

=108+1/4

=

(ii) a^b+b^a

Here, a=3 and b=−2.

Put the values in the expression a^b+b^a.

3⁻²+(−2)³

=(1/3)²+(−8)

=1/9 - 8

=1−72 / 9

=

(iii) (a+b)ab

Here, a=3 and b=−2.

Put the values in the expression (a+b)^ab.

[3+(−2)]³⁽⁻²⁾

=(1)⁻⁶

=1

Q.3. Prove that:

(i) 

(ii)

Proof: 

(i) 

Consider the left hand side:

=x^(a3−b3)×x^(b3−c3)×x^(c3−a3)

=x^{(a3−b3+b3−c3+c3−a3)}

=x0

=1

Left hand side is equal to right hand side.

Hence proved.

(ii)

Consider the left hand side:

= 1

Left hand side is equal to right hand side.

Hence proved.

Q.4. Prove that:

(i)

(ii)

Proof: (i) Consider the left hand side:

= 1

Therefore left hand side is equal to the right hand side. Hence proved.

(ii)

Consider the left hand side:

=1 

Therefore left hand side is equal to the right hand side. Hence proved.

Q.5. Prove that:

(i)= abc

(ii) (a⁻¹+b⁻¹)⁻¹=

Proof: (i) Consider the left hand side:

= abc

Therefore left hand side is equal to the right hand side. Hence proved.

(ii)

Consider the left hand side:

(a⁻¹+b⁻¹)⁻¹

Therefore left hand side is equal to the right hand side. Hence proved.

Q.6. If abc = 1, show that= 1

Proof: Consider the left hand side:

(abc=1) 

=1

Hence proved.

Q.7. Simplify the following:

(i)

(ii)

(iii)

(iv)

Proof: (i)

=3^3n+2−3n+3

=3⁵

=243 

(ii)

=4/24

=1/6

(iii)

= 19

(iv)

= 4

Q.8. Solve the following equations for x:

(i) 7^2x+3=1

(ii) 2^x+1=4^x−3

(iii) 2^5x+3=8^x+3

(iv) 4^2x=1/32

(v) 4^x−1×(0.5)^3−2x=(1/8)^x

(vi) 2^3x−7=256

Proof: (i)

7^2x+3=1

⇒7^2x+3=7⁰

⇒2^x+3=0

⇒2x=−3

⇒x=−3/2

(ii)

2^x+1=4^x−3

⇒2^x+1=(2²)^x−3

⇒2^x+1=(2^2x−6)

⇒x+1=2x−6

⇒x=7

(iii)

2^5x+3=8^x+3

⇒2^5x+3=(2³)^x+3

⇒2^5x+3=2^3x+9

⇒5x+3=3x+9

⇒2x=6

⇒x=3

(iv)

4^2x=1/32

⇒(2²)^2x=1/2⁵

⇒2^4x×2⁵=1

⇒2^4x+5=20

⇒4x+5=0

⇒x=−5/4

(v)

4^x−1×(0.5)^3−2x=(1/8)^x

⇒2^2x−2×2^2x−3=2^−3x

⇒2^{2x−2+2x−3}=2^−3x

⇒2^4x−5=2^−3x

⇒4x−5= −3x

⇒7x= 5

⇒x= 5/7

(vi)

2^3x−7=256

⇒2^3x−7=2⁸

⇒3x−7=8

⇒3x=15

⇒x=5

Q.9. Solve the following equations for x:

(i) 2^2x−2^x+3+2⁴=0

(ii) 3^2x+4+1=2.3^x+2

Proof: (i) 

2^2x−2^x+3+2⁴=0

⇒(2^x)²−(2^x×2³)+(2²)²=0

⇒(2^x)²−2×2^x×2²+(2²)²=0

⇒(2^x−2²)²=0

⇒2^x−2²=0

⇒2^x=2²

⇒x=2

(ii)

3^2x+4+1=2.3^x+2

⇒(3x+2)²−2.3^x+2+1=0

⇒(3^x+2−1)²=0

⇒3^x+2−1=0

⇒3^x+2=1=3⁰

⇒x+2=0

⇒x=−2

Q.10. If 49392=a⁴b²c³, find the values of a, b and c, where a, b and c are different positive primes.

Proof: First find out the prime factorization of 49392.

It can be observed that 49392 can be written as 2⁴×3²×7³, where 2, 3 and 7 are positive primes.

∴49392 = 2⁴3²7³ = a⁴b²c³

⇒a=2, b=3, c=7

Q.11. If 1176=2^a3^b7^c, find a, b and c.

Proof: First find out the prime factorization of 1176.

It can be observed that 1176 can be written as 2³×3¹×7².

1176 = 2³3¹7² = 2^a3^b7^c

Hence, a = 3, b = 1 and c = 2.

Q.12. Given 4725=3^a5^b7^c, find

(i) the integral values of a, b and c

(ii) the value of 2^−a3^b7^c

Proof: (i) Given 4725=3^a5^b7^c

First find out the prime factorization of 4725.

It can be observed that 4725 can be written as 3³×5²×7¹.

∴4725 = 3^a5^b7^c = 3³5²7¹

Hence, a = 3, b = 2 and c = 1.

(ii)

When a = 3, b = 2 and c = 1,

2^−a3^b7^c = 2⁻³×3²×7¹ = 1/8 × 9 × 7=63/8

Hence, the value of 2^−a3^b7^c is

Q.13. If a = xy^p−1,b = xy^q−1 and c = xy^r−1, prove that a^q−rb^r−pc^p−q = 1.

Proof: It is given that a = xy^p−1,b = xy^q−1 and c = xy^r−1

=x⁰y⁰

=1