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 x2 y z2
(vi)
Solution:
We have,
=
=
=
=
(vii)
Solution:
We have,
=
=
=
=
Question 2. Simplify
(i)
Solution:
We have,
=
=
=
=
(ii)
Solution:
We have,
=
=
= 2−3
=
(iii)
Solution:
We have,
=
=
= 7−2
=
(iv)
Solution:
We have,
=
=
=
= 0.1
(v)
Solution:
We have,
=
=
=
=
=
(vi)
Solution:
We have,
=
=
=
=
=
(vii)
Solution:
We have,
=
=
=
= 52 × 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 − 3 − 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 × 1 × 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. =
=
=
=
=
= 28√2
= 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
= [x0]a+b
= x0
= 1
= R.H.S.
Hence proved.
(iii)
Solution:
We have,
L.H.S. =
=
=
=
= x0
= 1
= R.H.S.
Hence proved.
(iv)
Solution:
We have,
L.H.S. =
=
=
=
= R.H.S.
Hence proved.
(v) (xa-b)a+b (xb-c)b+c (xc-a)c+a = 1
Solution:
We have,
L.H.S. = (xa-b)a+b (xb-c)b+c (xc-a)c+a
=
=
= x0
= 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. =
= (ax-y)x+y (ay-z)y+z (ax-z)x+z
=
=
= a0
= 1
= R.H.S.
Hence proved.
(viii)
Solution:
We have,
L.H.S. =
= (3a-b)a+b (3b-c)b+c (3c-a)c+a
=
=
= 30
= 1
= R.H.S.
Hence proved.
Question 5. If 2x = 3y = 12z, show that 1/z = 1/y + 2/x.
Solution:
We are given,
=> 2x = 3y = 12z = k (say)
So, we get,
=> 12 = k1/z
=> 2 × 3 × 2 = k1/z
=> 22 × 3 = k1/z
=> (k1/x)2 × (k)1/y = k1/z
=> (k)2/x × (k)1/y = k1/z
=> = k1/z
=> 2/x + 1/y = 1/z
Hence proved.
Question 6. If 2x = 3y = 6−z, show that 1/x + 1/y + 1/z = 0.
Solution:
We are given,
=> 2x = 3y = 6−z = k (say)
So, we get,
=> 6 = k-1/z
=> (2 × 3) = k-1/z
=> k1/x × k1/y = k-1/z
=> = k-1/z
=> 1/x + 1/y = −1/z
=> 1/x + 1/y + 1/z = 0
Hence proved.
Question 7. If ax = by = cz and b2 = ac, then show that y = 2zx/(z+x).
Solution:
We are given,
=> ax = by = cz = k (say)
=> a = k1/x, b = k1/y, c = k1/z
We are given, b2 = ac
=> (k1/y)2 = k1/x × k1/z
=> k2/y =
=> 2/y = 1/x + 1/z
=> 2/y = (x+z)/xz
=> y = 2zx/(z+x)
Hence proved.
Question 8. If 3x = 5y = (75)z, show that z = xy/(2x+y).
Solution:
We are given,
=> 3x = 5y = (75)z = k (say)
So, we get,
=> 75 = k1/z
=> 3 × 52 = k1/z
=> (k)1/x × (k1/y)2 = k1/z
=> (k)1/x × (k)2/y = k1/z
=> = k1/z
=> 1/x + 2/y = 1/z
=> (2x+y)/xy = 1/z
=> z = xy/(2x+y)
Hence proved.
Question 9. If (27)x = 9/3x, find x.
Solution:
We are given,
=> (27)x = 9/3x
=> (33)x = 32/3x
=> 33x = 32−x
=> 3x = 2 − x
=> 4x = 2
=> x = 2/4
=> x = 1/2
Question 10. Find the values of x in each of the following:
(i) 25x ÷ 2x =
Solution:
We have,
=> 25x ÷ 2x =
=> 25x−x =
=> 24x = 24
=> 4x = 4
=> x = 1
(ii) (23)4 = (22)x
Solution:
We have,
=> (23)4 = (22)x
=> 212 = 22x
=> 2x = 12
=> x = 6
(iii)
Solution:
We have,
=>
=>
=>
=>
=>
=> x = 3
(iv) 5x−2 × 32x−3 = 135
Solution:
We have,
=> 5x−2 × 32x−3 = 135
=> 5x−2 × 32x−3 = 5 × 27
=> 5x−2 × 32x−3 = 51 × 33
=> x − 2 = 1 and 2x − 3 = 3
=> x = 3
(v) 2x−7 × 5x−4 = 1250
Solution:
We are given,
=> 2x−7 × 5x−4 = 1250
=> 2x−7 × 5x−4 = 2 × 625
=> 2x−7 × 5x−4 = 2 × 54
=> x − 7 = 1 and x − 4 = 4
=> x = 8
(vi)
Solution:
We have,
=>
=>
=>
=> 4x/3 + 1/3 = −5
=> 4x +1 = −15
=> 4x = −16
=> x = −4
(vii) 52x+3 = 1
Solution:
We have,
=> 52x+3 = 1
=> 52x+3 = 50
=> 2x + 3 = 0
=> 2x = −3
=> x = −3/2
(viii)
Solution:
We have,
=>
=> = 256 − 81 − 6
=> = 169
=>
=> √x = 2
=> x = 4
(ix)
Solution:
We have,
=>
=>
=>
=> (x+1)/2 = −3
=> x + 1 = −6
=> x = −7
Question 11. If x = 21/3 + 22/3, show that x3 − 6x = 6.
Solution:
Given, x = 21/3 + 22/3
Therefore, x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(21/3 + 22/3)
=> x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(x)
=> x3 = 2 + 4 + 3(2)(x)
=> x3 = 6 + 6x
=> x3 − 6x = 6
Hence proved.
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