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

**Question 12. Determine (8x)**^{x}, if 9^{x+2} = 240 + 9^{x}.

^{x}, if 9

^{x+2}= 240 + 9

^{x}.

**Solution:**

We have,

=> 9

^{x+2}= 240 + 9^{x}=> 9

^{x+2}âˆ’ 9^{x}= 240=> 9

^{x}(9^{2 }âˆ’ 1) = 240=> 9

^{x}= 240/80=> 3

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

=> x = 1/2

Therefore, (8x)

^{x}= [8 Ã— (1/2)]^{1/2}= 4

^{1/2}= 2

**Question 13. If 3**^{x+1} = 9^{xâˆ’2}, find the value of 2^{1+x}.

^{x+1}= 9

^{xâˆ’2}, find the value of 2

^{1+x}.

**Solution:**

We have,

=> 3

^{x+1}= 9^{xâˆ’2}=> 3

^{x+1}= (3^{2})^{xâˆ’2}=> 3

^{x+1}= 3^{2xâˆ’4}=> x + 1 = 2x âˆ’ 4

=> x = 5

Therefore, 2

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

^{6}= 64

**Question 14. If 3**^{4x }= (81)^{âˆ’1} and (10)^{1/y} = 0.0001, find the value of 2^{âˆ’x+4y}.

^{4x }= (81)

^{âˆ’1}and (10)

^{1/y}= 0.0001, find the value of 2

^{âˆ’x+4y}.

**Solution:**

We are given,

=> 3

^{4x}= (81)^{âˆ’1}=> 3

^{4x}= (3^{4})^{âˆ’1}=> 3

^{4x}= (3)^{âˆ’4}=> 4x = âˆ’4

=> x = âˆ’1

And also, (10)

^{1/y }= 0.0001=> (10)

^{1/y}= (10)^{âˆ’4}=> 1/y = âˆ’4

=> y = âˆ’1/4

Therefore, 2

^{âˆ’x+4y }= 2^{1+4(âˆ’1/4) }= 2

^{1âˆ’1 }= 1

**Question 15. If 5**^{3x} = 125 and 10^{y} = 0.001. Find x and y.

^{3x}= 125 and 10

^{y}= 0.001. Find x and y.

**Solution:**

We are given,

=> 5

^{3x}= 125=> 5

^{3x}= 5^{3}=> 3x = 3

=> x =1

Also, (10)

^{y}= 0.001=> 10

^{y}= 10^{âˆ’3}=> y = âˆ’3

Therefore, the value of x is 1 and the value of y is â€“3.

**Question 16. Solve the following equations:**

**(i) 3**^{x+1} = 27 Ã— 3^{4}

^{x+1}= 27 Ã— 3

^{4}

**Solution:**

We have,

=> 3

^{x+1}= 27 Ã— 3^{4}=> 3

^{x+1}= 3^{3}Ã— 3^{4}=> 3

^{x+1 }= 3^{7}=> x + 1 = 7

=> x = 6

**(ii) **** **

**Solution:**

We have,

=>

=>

=>

=>

=> 4x = âˆ’8/y = 3

=> x = 3/4 and y = âˆ’8/3

**(iii) 3**^{xâˆ’1} Ã— 5^{2yâˆ’3} = 225

^{xâˆ’1}Ã— 5

^{2yâˆ’3}= 225

**Solution:**

We have,

=> 3

^{xâˆ’1}Ã— 5^{2yâˆ’3}= 225=> 3

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

=> x = 3 and 2y = 5

=> x = 3 and y = 5/2

**(iv) 8**^{x+1} = 16^{y+2} and (1/2)^{3+x} = (1/4)^{3y}

^{x+1}= 16

^{y+2}and (1/2)

^{3+x}= (1/4)

^{3y}

**Solution:**

We have,

=> 8

^{x+1}= 16^{y+2}=> (2

^{3})^{x+1}= (2^{4})^{y+2}=> 2

^{3x+3}= 2^{4y+8}=> 3x + 3 = 4y + 8 . . . . (1)

Also, (1/2)

^{3+x}= (1/4)^{3y}=> (1/2)

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

^{3+x}= (1/2)^{6y}=> 3 + x = 6y

=> x = 6y âˆ’ 3 . . . . (2)

Putting (2) in (1), we get,

=> 3(6y âˆ’ 3) + 3 = 4y + 8

=> 18y âˆ’ 9 + 3 = 4y + 8

=> 14y = 14

=> y = 1

Putting y = 1 in (2), we get,

x = 6(1) âˆ’ 3 = 6 âˆ’ 3 = 3

Therefore, the value of x is 1 and the value of y is â€“3.

**(v) 4**^{xâˆ’1} Ã— (0.5)^{3âˆ’2x} = (1/8)^{x}

^{xâˆ’1}Ã— (0.5)

^{3âˆ’2x}= (1/8)

^{x}

**Solution:**

We have,

=> 4

^{xâˆ’1}Ã— (0.5)^{3âˆ’2x}= (1/8)^{x}=> (2

^{2})^{xâˆ’1 }Ã— (1/2)^{3âˆ’2x}= [(1/2)^{3}]^{x}=> 2

^{2xâˆ’2}Ã— 2^{2xâˆ’3}= 2^{âˆ’3x}=> 2

^{2xâˆ’2+2xâˆ’3}= 2^{âˆ’3x}=> 2

^{4xâˆ’5}= 2^{âˆ’3x}=> 4x âˆ’ 5 = âˆ’3x

=> 7x = 5

=> x = 5/7

**(vi) **** **

**Solution:**

We have,

=>

=>

=> 1/2 = 2x âˆ’ 1

=> 2x = 3/2

=> x = 3/4

**Question: 17. If a and b are distinct positive primes such that, **** find x and y.**

**Solution:**

We have,

=>

=> (a

^{6 }b^{âˆ’4})^{1/3}= a^{x}b^{2y}=> a

^{6/3 }b^{âˆ’4/3}= a^{x}b^{2y}=> a

^{2 }b^{âˆ’4/3}= a^{x}b^{2y}=> x = 2 and 2y = âˆ’4/3

=> x = 2 and y = âˆ’2/3

**Question 18. If a and b are different positive primes such that,**

**(i) ****, find x and y.**

**Solution:**

We have,

=>

=> (a

^{âˆ’1âˆ’2 }b^{2+4})^{7}Ã· (a^{3+2 }b^{âˆ’5âˆ’3}) = a^{x}b^{y}=> (a

^{âˆ’3 }b^{6})^{7}Ã· (a^{5 }b^{âˆ’8}) = a^{x}b^{y}=> (a

^{âˆ’21 }b^{42}) Ã· (a^{5 }b^{âˆ’8}) = a^{x}b^{y}=> (a

^{âˆ’21âˆ’5 }b^{42+8}) = a^{x}b^{y}=> (a

^{âˆ’26 }b^{50}) = a^{x}b^{y}=> x = âˆ’26, y = 50

**(ii) (a + b)**^{âˆ’1}(a^{âˆ’1} + b^{âˆ’1}) = a^{x}b^{y}, find x+y+2.

^{âˆ’1}(a

^{âˆ’1}+ b

^{âˆ’1}) = a

^{x}b

^{y}, find x+y+2.

**Solution:**

We have,

=> (a + b)

^{âˆ’1}(a^{âˆ’1}+ b^{âˆ’1}) = a^{x}b^{y}=> = a

^{x}b^{y}=> = a

^{x}b^{y}=> 1/ab = a

^{x}b^{y}=> a

^{âˆ’1}b^{âˆ’1 }= a^{x}b^{y}=> x = âˆ’1 and y = âˆ’1

So, x+y+2 = âˆ’1âˆ’1+2 = 0.

**Question 19. If 2**^{x} Ã— 3^{y} Ã— 5^{z} = 2160, find x, y and z. Hence compute the value of 3^{x} Ã— 2^{âˆ’y} Ã— 5^{âˆ’z}.

^{x}Ã— 3

^{y}Ã— 5

^{z}= 2160, find x, y and z. Hence compute the value of 3

^{x}Ã— 2

^{âˆ’y}Ã— 5

^{âˆ’z}.

**Solution:**

We are given,

=> 2

^{x}Ã— 3^{y }Ã— 5^{z}= 2160=> 2

^{x }Ã— 3^{y}Ã— 5^{z}= 2^{4 }Ã— 3^{3 }Ã— 5^{1}=> x = 4, y = 3, z = 1

Therefore, 3

^{x}Ã— 2^{âˆ’y}Ã— 5^{âˆ’z}= 3^{4}Ã— 2^{âˆ’3}Ã— 5^{âˆ’1}= (81) (1/8) (1/5)

= 81/40

**Question 20. If 1176 = 2**^{a} Ã— 3^{b} Ã— 7^{c}, find the values of a, b and c. Hence, compute the value of 2^{a} Ã— 3^{b} Ã— 7^{-c} as a fraction.

^{a}Ã— 3

^{b}Ã— 7

^{c}, find the values of a, b and c. Hence, compute the value of 2

^{a}Ã— 3

^{b}Ã— 7

^{-c}as a fraction.

**Solution:**

We are given,

=> 1176 = 2

^{a}Ã— 3^{b}Ã— 7^{c}=> 2

^{3}Ã— 3^{1}Ã— 7^{2 }= 2a Ã— 3b Ã— 7c=> a = 3, b = 1, c = 2

Therefore, 2

^{a}Ã— 3^{b}Ã— 7^{âˆ’c}= 2^{3}Ã— 3^{1}Ã— 7^{âˆ’2}= (8) (3) (1/49)

= 24/49

**Question 21. Simplify**

**(i) **

**Solution:**

We have,

=

= (x

^{a+bâˆ’c})^{aâˆ’b }(x^{b+câˆ’a})^{bâˆ’c }(x^{c+aâˆ’b})^{câˆ’a}=

=

= x

^{0}= 1

**(ii) **

**Solution:**

We have,

=>

=>

=>

=>

=>

=>

=> x

^{0}= 1

**Question 22. Show that ****.**

**Solution:**

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

**Question 23. (i) If a = x**^{m+n}y^{l}, b = x^{n+l}y^{m} and c = x^{l+m}y^{n}, prove that a^{mâˆ’n }b^{nâˆ’l }c^{lâˆ’m} = 1.

^{m+n}y

^{l}, b = x

^{n+l}y

^{m}and c = x

^{l+m}y

^{n}, prove that a

^{mâˆ’n }b

^{nâˆ’l }c

^{lâˆ’m}= 1.

**Solution:**

Given, a = x

^{m+n}y^{l}, b = x^{n+l}y^{m}and c = x^{l+m}y^{n}.We have,

L.H.S. = a

^{mâˆ’n}b^{nâˆ’l}c^{lâˆ’m}= (x

^{m+n}y^{l})^{mâˆ’n}(x^{n+l}y^{m})^{nâˆ’l}(x^{l+m}y^{n})^{lâˆ’m}=

=

= x

^{0}y^{0}= 1

= R.H.S.

Hence proved.

**(ii) If x = a**^{m+n}, y = a^{n+l }and z = a^{l+m}, prove that x^{m}y^{n}z^{l} = x^{n}y^{l}z^{m}.

^{m+n}, y = a

^{n+l }and z = a

^{l+m}, prove that x

^{m}y

^{n}z

^{l}= x

^{n}y

^{l}z

^{m}.

**Solution:**

Given, x = a

^{m+n}, y = a^{n+l}and z = a^{l+m}.We have,

L.H.S. = x

^{m}y^{n}z^{l}= (a

^{m+n})^{m }(a^{n+l})^{n}(a^{l+m})^{l}=

=

= (a

^{m+n})^{n}(a^{n+l})^{l}(a^{l+m})^{m}= x

^{n}y^{l}z^{m}= R.H.S.

Hence proved.

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