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

  • Last Updated : 30 Apr, 2021

Question 12. Determine (8x)x, if 9x+2 = 240 + 9x.

Solution:

We have,

=> 9x+2 = 240 + 9x

=> 9x+2 − 9x = 240

=> 9x (92 − 1) = 240



=> 9x = 240/80

=> 32x = 3

=> 2x = 1

=> x = 1/2

Therefore, (8x)x = [8 × (1/2)]1/2

= 41/2

= 2

Question 13. If 3x+1 = 9x−2, find the value of 21+x.

Solution:

We have,

=> 3x+1 = 9x−2

=> 3x+1 = (32)x−2

=> 3x+1 = 32x−4

=> x + 1 = 2x − 4

=> x = 5

Therefore, 21+x = 21+5

= 26 

= 64

Question 14. If 34x = (81)−1 and (10)1/y = 0.0001, find the value of 2−x+4y.

Solution:



We are given,

=> 34x = (81)−1

=> 34x = (34)−1

=> 34x = (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 = 21+4(−1/4) 

= 21−1 

= 1

Question 15. If 53x = 125 and 10y = 0.001. Find x and y.

Solution:

We are given,

=> 53x = 125

=> 53x = 53

 => 3x = 3

=> x =1

Also, (10)y = 0.001



=> 10y = 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) 3x+1 = 27 × 34

Solution:

We have,

=>  3x+1 = 27 × 34

=> 3x+1 = 33 × 34

=> 3x+1 = 37

=> x + 1 = 7

=> x = 6

(ii) 4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2   

Solution:

We have,

=> 4^{2x}=\left(\sqrt[3]{16}\right)^{\frac{-6}{y}}=(\sqrt{8})^2

=> (2^2)^{2x}=\left(\sqrt[3]{2^4}\right)^{\frac{-6}{y}}=(\sqrt{2^3})^2

=> (2)^{4x}=\left(2^\frac{4}{3}\right)^{\frac{-6}{y}}=(2^\frac{3}{2})^2

=> (2)^{4x}=(2)^{\frac{-8}{y}}=2^3

=> 4x = −8/y = 3

=> x = 3/4 and y = −8/3

(iii) 3x−1 × 52y−3 = 225

Solution:

We have,



=> 3x−1 × 52y−3 = 225

=> 3x−1 × 52y−3 = 32 × 52

=> x − 1 = 2 and 2y − 3 = 2 

=> x = 3 and 2y = 5

=> x = 3 and y = 5/2

(iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y

Solution:

We have,

=> 8x+1 = 16y+2

=> (23)x+1 = (24)y+2

=> 23x+3 = 24y+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) 4x−1 × (0.5)3−2x = (1/8)x

Solution:

We have,

=> 4x−1 × (0.5)3−2x = (1/8)x

=> (22)x−1 × (1/2)3−2x = [(1/2)3]x

=> 22x−2 × 22x−3 = 2−3x

=> 22x−2+2x−3 = 2−3x

=> 24x−5 = 2−3x

=> 4x − 5 = −3x

=> 7x = 5

=> x = 5/7

(vi) \sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}   

Solution:

We have,

=> \sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2x}

=> \left(\frac{a}{b}\right)^{\frac{1}{2}}=\left(\frac{a}{b}\right)^{2x-1}

=> 1/2 = 2x − 1

=> 2x = 3/2



=> x = 3/4

Question: 17. If a and b are distinct positive primes such that, \sqrt[3]{a^6b^{-4}}=a^xb^{2y} find x and y.

Solution:

We have,

=> \sqrt[3]{a^6b^{-4}}=a^xb^{2y}

=> (a6 b−4)1/3 = axb2y

=> a6/3 b−4/3 = axb2y

=> a2 b−4/3 = axb2y

=> 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) \left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y  , find x and y.

Solution:



We have,

=> \left(\frac{a^{-1}b^{2}}{a^2b^{-4}}\right)^7÷\frac{a^{3}b^{-5}}{a^{-2}b^{3}}=a^xb^y   

=> (a−1−2 b2+4)7 ÷ (a3+2 b−5−3) = axby

=> (a−3 b6)7 ÷ (a5 b−8) = axby

=> (a−21 b42) ÷ (a5 b−8) = axby

=> (a−21−5 b42+8) = axby

=> (a−26 b50) = axby

=> x = −26, y = 50

(ii) (a + b)−1(a−1 + b−1) = axby, find x+y+2.

Solution:

We have,

=> (a + b)−1(a−1 + b−1) = axby

=> (\frac{1}{a+b})(\frac{1}{a}+\frac{1}{b})   = axby 

=> (\frac{1}{a+b})(\frac{b+a}{ab})   = axby 

=> 1/ab = axby

=> a−1b−1 = axby

=> x = −1 and y = −1

So, x+y+2 = −1−1+2 = 0.

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

Solution:

We are given,

=> 2x × 3y × 5z = 2160



=> 2x × 3y × 5z = 24 × 33 × 51

=> x = 4, y = 3, z = 1

Therefore, 3x × 2−y × 5−z = 34 × 2−3 × 5−1

= (81) (1/8) (1/5)

= 81/40

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

Solution:

We are given,

=> 1176 = 2a × 3b × 7c

=> 23 × 31 × 72 = 2a × 3b × 7c

=> a = 3, b = 1, c = 2 

Therefore, 2a × 3b × 7−c = 23 × 31 × 7−2

= (8) (3) (1/49)

= 24/49

Question 21. Simplify

(i) \left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}

Solution:

We have,

\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}

= (xa+b−c)a−b (xb+c−a)b−c (xc+a−b)c−a

(x^{a^2-b^2-ca+cb})(x^{b^2-c^2-ab+ac})(x^{c^2-a^2-bc+ab})

(x^{a^2-b^2-ca+cb+b^2-c^2-ab+ac+c^2-a^2-bc+ab})

= x0



= 1

(ii) \sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}

Solution:

We have,

=> \sqrt[lm]{\frac{x^l}{x^m}}×\sqrt[mn]{\frac{x^m}{x^n}}×\sqrt[nl]{\frac{x^n}{x^l}}   

=> (x^{l-m})^{\frac{1}{lm}}×(x^{m-n})^{\frac{1}{mn}}×(x^{n-l})^{\frac{1}{nl}}

=> x^{\frac{l-m}{lm}}×x^{\frac{m-n}{mn}}×x^{\frac{n-l}{nl}}

=> x^{\frac{l-m}{lm}+\frac{m-n}{mn}+\frac{n-l}{nl}}

=> x^{\frac{nl-mn+ml-nl+mn-ml}{mnl}}

=> x^{\frac{0}{mnl}}

=> x0



= 1

Question 22. Show that \frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}=(\frac{a}{b})^{m+n}  .

Solution:

We have,

L.H.S. = \frac{(a+\frac{1}{b})^m×(a-\frac{1}{b})^n}{(b+\frac{1}{a})^m×(b-\frac{1}{a})^n}

\frac{(\frac{ab+1}{b})^m×(\frac{ab-1}{b})^n}{(\frac{ab+1}{a})^m×(\frac{ab-1}{a})^n}

(\frac{a}{b})^m×(\frac{a}{b})^n

(\frac{a}{b})^{m+n}

= R.H.S.

Hence proved.

Question 23. (i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−n bn−l cl−m = 1.

Solution:

Given, a = xm+nyl, b = xn+lym and c = xl+myn

We have,

L.H.S. = am−n bn−l cl−m

= (xm+nyl)m−n(xn+lym)n−l(xl+myn)l−m

(x^{m^2-n^2}y^{lm-ln})(x^{n^2-l^2}y^{mn-ml})(x^{l^2-m^2}y^{nl-mn})

x^{m^2-n^2+n^2-l^2+l^2-m^2}y^{lm-ln+mn-ml+nl-mn}

= x0y0

= 1

= R.H.S.

Hence proved.



(ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm.

Solution:

Given, x = am+n, y = an+l and z = al+m.

We have,

L.H.S. = xmynzl

= (am+n)m (an+l)n (al+m)l

a^{m^2+mn+n^2+ln+l^2+ml}

a^{mn+n^2}×a^{nl+l^2}×a^{lm+m^2}

= (am+n)n (an+l)l (al+m)m

= xnylzm

= R.H.S.

Hence proved.

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