# 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}− 9^{x}= 240=> 9

^{x}(9^{2 }− 1) = 240=> 9

^{x}= 240/80=> 3

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

=> x = 1/2

Therefore, (8x)

^{x}= [8 × (1/2)]^{1/2}= 4

^{1/2}= 2

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

^{x+1}= 9

^{x−2}, find the value of 2

^{1+x}.

**Solution:**

We have,

=> 3

^{x+1}= 9^{x−2}=> 3

^{x+1}= (3^{2})^{x−2}=> 3

^{x+1}= 3^{2x−4}=> x + 1 = 2x − 4

=> x = 5

Therefore, 2

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

^{6}= 64

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

^{4x }= (81)

^{−1}and (10)

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

^{−x+4y}.

**Solution:**

We are given,

=> 3

^{4x}= (81)^{−1}=> 3

^{4x}= (3^{4})^{−1}=> 3

^{4x}= (3)^{−4}=> 4x = −4

=> x = −1

And also, (10)

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

^{1/y}= (10)^{−4}=> 1/y = −4

=> y = −1/4

Therefore, 2

^{−x+4y }= 2^{1+4(−1/4) }= 2

^{1−1 }= 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^{−3}=> y = −3

Therefore, the value of x is 1 and the value of y is –3.

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

**(i) 3**^{x+1} = 27 × 3^{4}

^{x+1}= 27 × 3

^{4}

**Solution:**

We have,

=> 3

^{x+1}= 27 × 3^{4}=> 3

^{x+1}= 3^{3}× 3^{4}=> 3

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

=> x = 6

**(ii) **** **

**Solution:**

We have,

=>

=>

=>

=>

=> 4x = −8/y = 3

=> x = 3/4 and y = −8/3

**(iii) 3**^{x−1} × 5^{2y−3} = 225

^{x−1}× 5

^{2y−3}= 225

**Solution:**

We have,

=> 3

^{x−1}× 5^{2y−3}= 225=> 3

^{x−1}× 5^{2y−3}= 3^{2}× 5^{2}=> x − 1 = 2 and 2y − 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 − 3 . . . . (2)

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

=> 3(6y − 3) + 3 = 4y + 8

=> 18y − 9 + 3 = 4y + 8

=> 14y = 14

=> y = 1

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

x = 6(1) − 3 = 6 − 3 = 3

Therefore, the value of x is 1 and the value of y is –3.

**(v) 4**^{x−1} × (0.5)^{3−2x} = (1/8)^{x}

^{x−1}× (0.5)

^{3−2x}= (1/8)

^{x}

**Solution:**

We have,

=> 4

^{x−1}× (0.5)^{3−2x}= (1/8)^{x}=> (2

^{2})^{x−1 }× (1/2)^{3−2x}= [(1/2)^{3}]^{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) **** **

**Solution:**

We have,

=>

=>

=> 1/2 = 2x − 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^{−4})^{1/3}= a^{x}b^{2y}=> a

^{6/3 }b^{−4/3}= a^{x}b^{2y}=> a

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

=> x = 2 and y = −2/3

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

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

**Solution:**

We have,

=>

=> (a

^{−1−2 }b^{2+4})^{7}÷ (a^{3+2 }b^{−5−3}) = a^{x}b^{y}=> (a

^{−3 }b^{6})^{7}÷ (a^{5 }b^{−8}) = a^{x}b^{y}=> (a

^{−21 }b^{42}) ÷ (a^{5 }b^{−8}) = a^{x}b^{y}=> (a

^{−21−5 }b^{42+8}) = a^{x}b^{y}=> (a

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

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

^{−1}(a

^{−1}+ b

^{−1}) = a

^{x}b

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

**Solution:**

We have,

=> (a + b)

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

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

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

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

^{−1}b^{−1 }= a^{x}b^{y}=> x = −1 and y = −1

So, x+y+2 = −1−1+2 = 0.

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

^{x}× 3

^{y}× 5

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

^{x}× 2

^{−y}× 5

^{−z}.

**Solution:**

We are given,

=> 2

^{x}× 3^{y }× 5^{z}= 2160=> 2

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

Therefore, 3

^{x}× 2^{−y}× 5^{−z}= 3^{4}× 2^{−3}× 5^{−1}= (81) (1/8) (1/5)

= 81/40

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

^{a}× 3

^{b}× 7

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

^{a}× 3

^{b}× 7

^{-c}as a fraction.

**Solution:**

We are given,

=> 1176 = 2

^{a}× 3^{b}× 7^{c}=> 2

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

Therefore, 2

^{a}× 3^{b}× 7^{−c}= 2^{3}× 3^{1}× 7^{−2}= (8) (3) (1/49)

= 24/49

**Question 21. Simplify**

**(i) **

**Solution:**

We have,

=

= (x

^{a+b−c})^{a−b }(x^{b+c−a})^{b−c }(x^{c+a−b})^{c−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^{m−n }b^{n−l }c^{l−m} = 1.

^{m+n}y

^{l}, b = x

^{n+l}y

^{m}and c = x

^{l+m}y

^{n}, prove that a

^{m−n }b

^{n−l }c

^{l−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

^{m−n}b^{n−l}c^{l−m}= (x

^{m+n}y^{l})^{m−n}(x^{n+l}y^{m})^{n−l}(x^{l+m}y^{n})^{l−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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