# Class 9 RD Sharma Solutions – Chapter 2 Exponents of Real Numbers- Exercise 2.2 | Set 1

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

**(i) **

**Solution:**

We have,

=

=

=

=

**(ii) **

**Solution:**

We have,

=

=

=

**(iii) **

**Solution:**

We have,

=

=

=

**(iv) **** **

**Solution:**

We have,

=

=

=

=

=

**(v) **

**Solution:**

We have,

=

=

=

= 3 x

^{2 }y z^{2}

**(vi) **

**Solution:**

We have,

=

=

=

=

**(vii) **

**Solution:**

We have,

=

=

=

=

**Question 2. Simplify**

**(i) **** **

**Solution:**

We have,

=

=

=

=

**(ii) **

**Solution:**

We have,

=

=

= 2

^{âˆ’3}=

**(iii) **

**Solution:**

We have,

=

=

= 7

^{âˆ’2}=

**(iv) **

**Solution:**

We have,

=

=

=

= 0.1

**(v) **

**Solution:**

We have,

=

=

=

=

=

**(vi) **** **

**Solution:**

We have,

=

=

=

=

=

**(vii) **

**Solution:**

We have,

=

=

=

= 5

^{2}Ã— 7= 175

**Question 3. Prove that**

**(i) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

**(ii) **

**Solution:**

We have,

L.H.S. =

=

=

= 27 âˆ’ 3 âˆ’ 9

= 15

= R.H.S.

Hence proved.

**(iii) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

**(iv) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= 2 Ã— 1 Ã— 5

= 10

= R.H.S.

Hence proved.

**(v) **

**Solution:**

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

**(vi) **

**Solution:**

We have,

L.H.S. =

=

= 1 +

=

= R.H.S.

Hence proved.

**(vii) **

**Solution:**

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

**(viii) **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= 28âˆš2

= R.H.S.

Hence proved.

**(ix) **** **

**Solution:**

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

**Question 4. Show that**

**(i) **

**Solution:**

We have,

L.H.S. =

=

=

=

= 1

= R.H.S.

Hence proved.

**(ii) **

**Solution:**

We have,

L.H.S. =

=

=

= [x

^{-2ab-(-2ab)}]^{a+b}= [x

^{0}]^{a+b}= x

^{0}= 1

= R.H.S.

Hence proved.

**(iii) **

**Solution:**

We have,

L.H.S. =

=

=

=

= x

^{0}= 1

= R.H.S.

Hence proved.

**(iv) **

**Solution:**

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

**(v) (x**^{a-b})^{a+b} (x^{b-c})^{b+c} (x^{c-a})^{c+a} = 1

^{a-b})

^{a+b}(x

^{b-c})

^{b+c}(x

^{c-a})

^{c+a}= 1

**Solution:**

We have,

L.H.S. = (x

^{a-b})^{a+b}(x^{b-c})^{b+c}(x^{c-a)c+a }=

=

= x

^{0}= 1

= R.H.S.

Hence proved.

**(vi) **

**Solution:**

We have,

L.H.S. =

=

=

= x

= R.H.S.

Hence proved.

**(vii) **

**Solution:**

We have,

L.H.S. =

= (a

^{x-y})^{x+y}(a^{y-z})^{y+z}(a^{x-z})^{x+z }=

=

= a

^{0}= 1

= R.H.S.

Hence proved.

**(viii) **

**Solution:**

We have,

L.H.S. =

= (3

^{a-b})^{a+b}(3^{b-c})^{b+c}(3^{c-a})^{c+a }=

=

= 3

^{0}= 1

= R.H.S.

Hence proved.

**Question 5. If 2**^{x} = 3^{y} = 12^{z}, show that 1/z = 1/y + 2/x.

^{x}= 3

^{y}= 12

^{z}, show that 1/z = 1/y + 2/x.

**Solution:**

We are given,

=> 2

^{x}= 3^{y}= 12^{z}= k (say)So, we get,

=> 12 = k

^{1/z}=> 2 Ã— 3 Ã— 2 = k

^{1/z}=> 2

^{2}Ã— 3 = k^{1/z}=> (k

^{1/x})^{2}Ã— (k)^{1/y}= k^{1/z}=> (k)

^{2/x}Ã— (k)^{1/y}= k^{1/z}=> = k

^{1/z}=> 2/x + 1/y = 1/z

Hence proved.

**Question 6. If 2**^{x} = 3^{y} = 6^{âˆ’z}, show that 1/x + 1/y + 1/z = 0.

^{x}= 3

^{y}= 6

^{âˆ’z}, show that 1/x + 1/y + 1/z = 0.

**Solution:**

We are given,

=> 2

^{x}= 3^{y}= 6^{âˆ’z}= k (say)So, we get,

=> 6 = k

^{-1/z}=> (2 Ã— 3) = k

^{-1/z}=> k

^{1/x}Ã— k^{1/y}= k^{-1/z}=> = k

^{-1/z}=> 1/x + 1/y = âˆ’1/z

=> 1/x + 1/y + 1/z = 0

Hence proved.

**Question 7. If a**^{x} = b^{y} = c^{z} and b^{2} = ac, then show that y = 2zx/(z+x).

^{x}= b

^{y}= c

^{z}and b

^{2}= ac, then show that y = 2zx/(z+x).

**Solution:**

We are given,

=> a

^{x}= b^{y}= c^{z}= k (say)=> a = k

^{1/x}, b = k^{1/y}, c = k^{1/z}We are given, b

^{2}= ac=> (k

^{1/y})^{2}= k^{1/x}Ã— k^{1/z}=> k

^{2/y}==> 2/y = 1/x + 1/z

=> 2/y = (x+z)/xz

=> y = 2zx/(z+x)

Hence proved.

**Question 8. If 3**^{x }= 5^{y} = (75)^{z}, show that z = xy/(2x+y).

^{x }= 5

^{y}= (75)

^{z}, show that z = xy/(2x+y).

**Solution:**

We are given,

=> 3

^{x}= 5^{y}= (75)^{z}= k (say)So, we get,

=> 75 = k

^{1/z}=> 3 Ã— 5

^{2}= k^{1/z}=> (k)

^{1/x}Ã— (k^{1/y})^{2}= k^{1/z}=> (k)

^{1/x }Ã— (k)^{2/y}= k^{1/z}=> = k

^{1/z}=> 1/x + 2/y = 1/z

=> (2x+y)/xy = 1/z

=> z = xy/(2x+y)

Hence proved.

**Question 9. If (27)**^{x} = 9/3^{x}, find x.

^{x}= 9/3

^{x}, find x.

**Solution:**

We are given,

=> (27)

^{x}= 9/3^{x}=> (3

^{3})^{x}= 3^{2}/3^{x}=> 3

^{3x}= 3^{2âˆ’x}=> 3x = 2 âˆ’ x

=> 4x = 2

=> x = 2/4

=> x = 1/2

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

**(i) 2**^{5x} Ã· 2^{x} =

^{5x}Ã· 2

^{x}=

**Solution:**

We have,

=> 2

^{5x}Ã· 2^{x }==> 2

^{5xâˆ’x}==> 2

^{4x}= 2^{4}=> 4x = 4

=> x = 1

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

^{3})

^{4 }= (2

^{2})

^{x}

**Solution:**

We have,

=> (2

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

^{12}= 2^{2x}=> 2x = 12

=> x = 6

**(iii) **

**Solution:**

We have,

=>

=>

=>

=>

=>

=> x = 3

**(iv) 5**^{xâˆ’2 }Ã— 3^{2xâˆ’3} = 135

^{xâˆ’2 }Ã— 3

^{2xâˆ’3}= 135

**Solution:**

We have,

=> 5

^{xâˆ’2}Ã— 3^{2xâˆ’3}= 135=> 5

^{xâˆ’2}Ã— 3^{2xâˆ’3}= 5 Ã— 27=> 5

^{xâˆ’2 }Ã— 3^{2xâˆ’3}= 5^{1}Ã— 3^{3}=> x âˆ’ 2 = 1 and 2x âˆ’ 3 = 3

=> x = 3

**(v) 2**^{xâˆ’7 }Ã— 5^{xâˆ’4} = 1250

^{xâˆ’7 }Ã— 5

^{xâˆ’4}= 1250

**Solution:**

We are given,

=> 2

^{xâˆ’7}Ã— 5^{xâˆ’4}= 1250=> 2

^{xâˆ’7 }Ã— 5^{xâˆ’4 }= 2 Ã— 625=> 2

^{xâˆ’7}Ã— 5^{xâˆ’4}= 2 Ã— 5^{4}=> x âˆ’ 7 = 1 and x âˆ’ 4 = 4

=> x = 8

**(vi)**

**Solution:**

We have,

=>

=>

=>

=> 4x/3 + 1/3 = âˆ’5

=> 4x +1 = âˆ’15

=> 4x = âˆ’16

=> x = âˆ’4

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

^{2x+3}= 1

**Solution:**

We have,

=> 5

^{2x+3}= 1=> 5

^{2x+3}= 5^{0}=> 2x + 3 = 0

=> 2x = âˆ’3

=> x = âˆ’3/2

**(viii) **

**Solution:**

We have,

=>

=> = 256 âˆ’ 81 âˆ’ 6

=> = 169

=>

=> âˆšx = 2

=> x = 4

**(ix)**

**Solution:**

We have,

=>

=>

=>

=> (x+1)/2 = âˆ’3

=> x + 1 = âˆ’6

=> x = âˆ’7

**Question 11. If x = 2**^{1/3} + 2^{2/3}, show that x^{3} âˆ’ 6x = 6.

^{1/3}+ 2

^{2/3}, show that x

^{3}âˆ’ 6x = 6.

**Solution:**

Given, x = 2

^{1/3}+ 2^{2/3}Therefore, x

^{3}= (2^{1/3})^{3}+ (2^{2/3})^{3 }+ 3(2^{1/3})(2^{2/3})(2^{1/3}+ 2^{2/3})=> x

^{3 }= (2^{1/3})3 + (2^{2/3})^{3}+ 3(2^{1/3})(2^{2/3})(x)=> x

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

^{3}= 6 + 6x=> x

^{3}âˆ’ 6x = 6

Hence proved.

## Please

Loginto comment...