Exponents of Real Numbers- 2 RD Sharma Solutions | Mathematics (Maths) Class 9 PDF Download

RD Sharma Solutions Exercise 2.2 Exponents Of Real Numbers

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

(i) (√x⁻³)⁵

(ii) √x³ y⁻²

(iii) (x^−2/3y^−1/2)²

(iv) (√x)^−2/3√y⁴ ÷ √xy^−1/2

(v) 5√243 x¹⁰y⁵z¹⁰

(vi)

(vii)

Proof: We have to simplify the following, assuming thatare positive real numbers

(i) Given

As x is positive real number then we have

Hence the simplified value ofis

(ii) Given

By using law of rational exponentswe have

Hence the simplified value ofis

(iii) Given

As x and y are positive real numbers then we have 

By using law of rational exponentswe have

By using law of rational exponentswe have

Hence the simplified value ofis

(iv)

by using the law of rational exponents, a^m ÷ aⁿ = a^m−n, we have

(v)

=(243 × x¹⁰ × y⁵ × z¹⁰)^1/5

=(243)^1/5 × (x¹⁰)^1/5 × (y⁵)^1/5 × (z¹⁰)^1/5

=(3⁵)^1/5 × x^10×1/5 × y5^×1/5 × z^10×1/5

=3 × x² × y × z²

=3x²yz²

^(vi)

^(vii)

Q.2. Simplify:

(i)

(ii)

(iii) 

(v)

(v)

(vi)

(vii)

Proof: (1) Given

By using law of rational exponentswe have

Hence the value ofis

(ii)

(iii) Given

Hence the value ofis

(iv) Given

The value ofis

(v) Given

(vi) GivenSo,

By using the law of rational exponents

Hence the value ofis

(vii) GivenSo,

Hence the value ofis

Q.3. Prove that:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Proof: (i) We have to prove that

By using rational exponentwe get,

Hence,

(ii) We have to prove that

Hence,

iii) We have to prove that

Now,

Hence,

(iv) We have to prove that

Let

Hence,

(v) We have to prove that

Let

Hence,

(vi) We have to prove thatSo,

Let

Hence,

(vii) We have to prove that So let

Hence,

(viii) We have to prove thatSo,

Let

Hence,

(ix) We have to prove that

Let

Hence,

Q.4.  Show that:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Proof: (i)

= 1

(ii)

= 1

(iii)

(iv)

=x^2(a3+b3+c3)

(v)

= x⁰

= 1

(vi)

= x¹

= x

(vii)

=a⁰

=1

(viii)

=3⁰

=1

Q.5. If 27x =9/3^x,  find x

Proof: We are givenWe have to find the value of x

Since

y using the law of exponents we get,

On equating the exponents we get,

Hence,

Q.6. Find the values of x in each of the following:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii) 52^x+3=1

(viii)

(ix)

Proof: From the following we have to find the value of x

(i) Given

By using rational exponents

On equating the exponents we get,

The value of x is

(ii) Given

On equating the exponents

Hence the value of x is

(iii) Given

Comparing exponents we have,

Hence the value of x is

(iv) Given

On equating the exponents of 5 and 3 we get,

And,

The value of x is

(v) Given

On equating the exponents we get 

And, 

Hence the value of x is

(vi)

On comparing we get, 

⇒4x+1 = −15

⇒4x = −16

⇒x = −4

(vii) 5^2x+3=1

5^2x+3 =5⁰

⇒2^x+3 = 0

⇒x = −3/2

(viii) (13)^√x = 4⁴ − 3⁴ − 6

On comparing we get, 

√x = 2

on squaring both sides we get, 

x = 4

(ix)

On comparing we get, 

⇒x + 1 = −6

⇒x = −7

Q.7. If 3^4x = (81)⁻¹ and 10^1/y = 0.0001, find the value of 2^−x+4y.

Proof: It is given that 3^4x=(81)⁻¹ and 10^1/y=0.0001.

Now,

3^4x = (81)⁻¹

⇒3^4x = (3⁴)⁻¹

⇒(3^x)⁴ = (3⁻¹)⁴

⇒x = −1

And,

10^1/y=0.0001

⇒10^1/y = (1/10)⁴

⇒101/y = (10)⁻⁴

⇒1/y = −4

⇒y = −1/4

Therefore, the value of 2^−x+4y is 2^1+4(−1/4) = 2⁰ = 1

Q.8. If 5^3x=125 and 10^y=0.001, find x and y.

Proof: It is given that 5^3x = 125 and 10^y = 0.001.

Now,

5^3x = 125

⇒5^3x = 5³

⇒3x = 3

⇒x = 1

And,

10y=0.001

⇒10^y = 1/1000

⇒10^y = 10⁻³ 

⇒y = −3

Hence, the values of x and y are 1 and −3, respectively.

Q.9. Solve the following equations:

(i) 3^x+1 = 27×3⁴

(ii)

(iii) 3^x−1 × 5^2y−3 = 225

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

(v) 4^x−1 × (0.5)^3−2x = (178)^x

^{(vi)where a and b are distinct primes}

Proof: (i)

3^x+1 = 27×3⁴

⇒3^x+1 = 3³×3⁴

⇒3^x+1 = 3³⁺⁴

⇒3^x+1 = 3⁷

⇒x+1 = 7

⇒x = 6

(ii)

(iii) 

3^x−1 × 52^y−3 = 225

⇒3^x−1 × 5^2y−3 = 3 × 3 × 5 × 5

⇒3^x−1 × 5^2y−3 = 3² × 5²

⇒x − 1 = 2 and 2y − 3 = 2

⇒ x = 3 and y = 5/2

(iv)

⇒3x+3 = 4y+8 and 3+x = 6y

Now,

3+x = 6y ⇒ x = 6y − 3

Putting x = 6y − 3 in 3x − 4y = 5, we get

3(6y−3)−4y = 5

⇒18y − 9 − 4y = 5

⇒14y = 14

⇒y = 1

Putting y = 1 in x=6y−3, we get

x=6×1−3=3

(v)

⇒4x − 5 = −3x

⇒7x = 5

⇒x = 5/7

(vi)

⇒1/2 = 2x − 1

⇒3/2 = 2x

⇒x = 3/4

Q.10. If a and b are distinct primes such that= a^xb^2y, find x and y.

Proof: Given: = a^xb^2y

Q.11. If a and b are different positive primes such that

(i)= axby, find x and y.

(ii)= axby, find x + y + 2.

Proof: (i)

⇒a−²¹⁻⁵b⁴²⁺⁸ = axby

⇒a⁻²⁶b⁵⁰ = axby

⇒x = −26  and  y = 50

(ii)

⇒(ab)⁻¹ = a^xb^y

⇒a⁻¹b⁻¹ = a^xb^y

⇒x=−1  and  y=−1

Therefore, the value of x + y + 2 is −1 −1 + 2 = 0.

Q.12. If 2^x × 3^y × 5^z = 2160, find x , y and z. Hence, compute the value of 3^x × 2^{− y} × 5^{− z}.

Proof: Given: 2^x × 3^y × 5^z = 2160

First, find out the prime factorization of 2160.

It can be observed that 2160 can be written as 2⁴ × 3³ × 5¹.

Also,

2^x × 3^y × 5^z = 2⁴ × 3³ × 5¹

⇒x = 4, y = 3, z = 1

Therefore, the value of 3^x × 2^−y × 5^−z is 3⁴ × 2⁻³ × 5⁻¹ = 81

Q.13. 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.

Proof: First find the prime factorisation of 1176.

It can be observed that 1176 can be written as 2³ × 3¹ × 7².

1176 = 2^a3^b7^c = 2³3¹7²

So, a = 3, b = 1 and c = 2.

Therefore, the value of 2^a × 3^b × 7^−c  is 2³ × 3¹ × 7⁻²= 8 x 3 x 1/49 =

Q.14. Simplify:

(i)

(ii)

Proof: (i)

= 1

(ii)

= x⁰

= 1

Q.15. Show that:

Proof:  

Q.16. (i) If a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ, prove that a^m−nb^n−lc^l−m = 1.

(ii) If x = a^m+n, y = a^n+l and z = a^l+m, prove that x^myⁿz^l=xⁿy^lz^m.

Proof: (i) Given: a = x^m+ny^l, b = x^n+ly^m and c = x^l+myⁿ

= x⁰y⁰

= 1

(ii) Given:x = a^m+n, y = a^n+l and z = a^l+m

Putting the values of x, y and z in x^myⁿz^l, we get

x^myⁿz^l

=a^{m2+n2+l2+nm+ln+lm}

Putting the values of x, y and z in xⁿy^lz^m, we get

xⁿy^lz^m

So, x^myⁿz^l = xⁿy^lz^m

Multiple Choice Questions(MCQs)

Q.1. The value ofis

(a) 5

(b) 125

(c) 1/5

(d) -125

Proof: We have to find the value of

So,

The value ofis 125

Hence the correct choice is

Q.2. The value of x − y^x-y when x = 2 and y = −2 is

(a) 18

(b) −18

(c) 14

(d) −14

Proof: Given

Here x = 2, y = -2

By substituting in we get

The value ofis -14

Hence the correct choice is

Q.3. The product of the square root of x with the cube root of x is

(a) cube root of the square root of x

(b) sixth root of the fifth power of x

(c) fifth root of the sixth power of x

(d) sixth root of x

Proof: We have to find the product (say L) of the square root of x with the cube root of x is. So, 

The product of the square root of x with the cube root of x is

Hence the correct alternative is

Q.4. The seventh root of x divided by the eighth root of x is

(a) x

(b) √x

(c) 

(d) 

Proof: We have to find he seventh root of x divided by the eighth root of x, so let it be L. So, 

The seventh root of x divided by the eighth root of x is

Hence the correct choice is

Q.5. The square root of 64 divided by the cube root of 64 is

(a) 64

(b) 2

(c) 1/2

(d) 64^2/3

Proof: We have to find the value of

So,

The value ofis 2

Hence the correct choice is

Q.6. Which of the following is (are) not equal to

(a) 

(b)

(c)

(d)

Proof: We have to find the value of

So,

Hence the correct choice is

Q.7. When simplified (x⁻¹+y⁻¹)⁻¹ is equal to

(a) xy

(b) x+y

(c) xy/x+y

(d) x+y/xy

Proof: We have to simplify

So,

The value ofis

Hence the correct choice is

Q.8. If 8^x+1 = 64 , what is the value of 3^2x+1 ?

(a) 1

(b) 3

(c) 9

(d) 27

Proof: We have to find the value ofprovided

So,

Equating the exponents we get

By substitute inwe get 

The real value ofis 27

Hence the correct choice is

Q.9. If (2³)² = 4^x, then 3^x =

(a) 3

(b) 6

(c) 9

(d) 27

Proof: We have to find the value of 3^x provided(2³)² = 4^x

So,

2^3x2 = 2^2x

2⁶ = 22^x

By equating the exponents we get

By substituting in 3^x we get

The value of 3^x is 27

Hence the correct choice is

Q.10. If x^-2 = 64, then x^1/3+x⁰ =

(a) 2

(b) 3

(c) 3/2

(d) 2/3

Proof: We have to find the value of if

Consider,

Multiply on both sides of powers we get

By taking reciprocal on both sides we get,

Substitutinginwe get,

By taking least common multiply we get

Hence the correct choice is

Q.11. When simplifiedis

(a) 9

(b) −9

(c) 1/9

(d) −1/9

Proof: We have to find the value of

So,

Hence the correct choice is

Q.12. Which one of the following is not equal to?

(a)

(b)

(c)

(d)

Proof: We have to find the value of

So,

Also,

Hence the correct alternative is

Q.13. Which one of the following is not equal to?

(a)

(b)

(c)

(d)

Proof: We have to find the value of

So,

Since,is equal to ,,

Hence the correct choice is

Q.14. If a, b, c are positive real numbers, then is equal to 

(a) 1

(b) abc

(c) √abc

(d) 1/abc

Proof: We have to find the value of when a, b, c are positive real numbers.

So,

Taking square root as common we get

Hence the correct alternative is

Q.15. , then x =

(a) 2

(b) 3

(c) 4

(d) 1

Proof: We have to find value of x provided

So,

Equating exponents of power we get x = 4

Hence the correct alternative is

Q.16. The value of is

(a) 1/2

(b) 2

(c) 1/4

(d) 4

Proof: Find the value of

Hence the correct choice is

Q.17. If a, b, c are positive real numbers, thenis equal to

(a) 5a²bc²

(b) 25ab²c

(c) 5a³bc³

(d) 125a²bc²

Proof: Find value of

Hence the correct choice is

Q.18. If a, m, n are positive integers, thenis equal to

(a) a^mn

(b) a

(c) a^m/n

(d) 1

Proof: Find the value of

So,

Hence the correct choice is

Q.19. If x = 2 and y = 4, then=

(a) 4

(b) 8

(c) 12

(d) 2

Proof: We have to find the value of

Substitute x = 2, y = 4 into get

Hence the correct choice is

Q.20. The value of m for which=7^m,  is

(a) −1/3

(b) 1/4

(c) −3

(d) 2

Proof:  We have to find the value of for

By using rational exponents

7^−1/3=7^m

Equating power of exponents we get

Hence the correct choice is

Q.21. The value of, is

(a) 196

(b) 289

(c) 324

(d) 400

Proof: We have to find the value of

By using the identity we get,

Hence correct choice is

Q.22. (256)^0.16 × (256)^0.09

(a) 4

(b) 16

(c) 64

(d) 256.25

Proof: We have to find the value of

By using law of rational exponents

The value ofis 4

Hence the correct choice is

Q.23. If , then 10^-y equals

(a) −1/5

(b) 1/50

(c) 1/625

(d) 1/5

Proof: We have to find the value of 10^-y

Given that 10^2y = 25 therefore,

Hence the correct option is.

Q.24. If 9^x+2 = 240 + 9^x, then x  = z

(a) 0.5

(b) 0.2

(c) 0.4

(d) 0.1

Proof: We have to find the value of x

Given, 9^x+2 = 240 + 9^x

3^2x = 3¹

By equating the exponents we get 

2x = 1

x = 1/2

x = 0.5

Hence the correct alternative is.

Q.25. If x is a positive real number and x² = 2, then x³ =

(a) √2

(b) 2√2

(c) 3√2

(d) 4

Proof: We have to find x³ provided x² = 2. So,

By raising both sides to the power 1/2

By substitutingin x³ we get

The value of x³ is

Hence the correct choice is.

Q.26. Ifand x > 0, then x =

(a) √2/4

(b) 2√2

(c) 4

(d) 64  

Proof: For, we have to find the value of x

So,

By raising both sides to the power 2 we get

The value of x  is 64

Hence the correct alternative is

Q.27. If g = t^2/3 + 4^{t − 1/2}, What is the value of g when t = 64?

(a) 21/2

(b) 33/2

(c) 16

(d) 257/16

Proof: GivenWe have to find the value of 

So,

The value of g is 33/2

Hence the correct choice is

Q.28. If 4x − 4x⁻¹ = 24, then (2x)^x equals

(a) 5√5

(b) √5

(c) 25√5

(d) 125

Proof: We have to find the value of

So,

Taking 4x as common factor we get 

By equating powers of exponents we get

By substitutingin we get

Hence the correct choice is

Q.29. When simplifiedis

(a) 8

(b) 1/8

(c) 2

(d) 1/2

Proof:

Hence the correct choice is.

Q.30. If,  then x =

(a) 2

(b) 3

(c) 5

(d) 4

Proof: We have to find the value of x provided

So,

By cross multiplication we get

By equating exponents we get 

And,

Hence the correct choice is

Q.31. The value of 64^-1/3 (64^1/3-64^2/3), is

(a) 1

(b) 1/3

(c) −3

(d) −2

Proof: Find the value of

So,

Hence the correct statement is.

Q.32. If √5ⁿ = 125, then =

(a) 25

(b) 1/125

(c) 625

(d) 1/5

Proof: We have to findprovided

So,

Substitute n = 6  in to get

Hence the value ofis 25

The correct choice is

Q.33. If (16)^2x+3 =(64)^x+3, then 4^2x-2 =

(a) 64

(b) 256

(c) 32

(d) 512

Proof: We have to find the value of 4^2x-2 provided

So,

Equating the power of exponents we get

x = 3

The value of 4^2x-2 is

Hence the correct alternative is

Q.34. If thenis equal to

(a) 1/2

(b) 2  

(c) 4

(d) −1/4

Proof: We have to find the value ofprovided

Consider,

Equating the power of exponents we get 

2m = 2

m = 2/2

m = 1

By substituting we get 

Hence the correct choice is.

Q.35. If,and a = 2^1/10 , then=

a) 2

(b) 1/4

(c) 9

(d) 1/8

Proof: Given :, and a = 2^1/10

To find :

Find :

By using rational components We get

By equating rational exponents we get

Now,= (a^2m+n−p).(a^m−2n+2p) we get

=a^{2m+n−p+m−2n+2p}

=a^3m−n+p

Now putting value of a = 2^1/10 we get,

On comparing LHS and RHS we get, p - n = 4.

Now, 

= a^{3m - n + p}

= 2

So, option (a) is the correct answer.

Q.36. If=3⁷ , then x 

(a) 3

(b) −3

(c) 1/3

(d) −1/3

Proof:  We have to find the value of x provided

So,

By using law of rational exponents we get

By equating exponents we get

Hence the correct choice is.

Q.37. If o <y <x, which statement must be true?

(a) √x−√y = √x−y

(b) √x + √x = √2x

(c) x√y = y√x

(d) √xy = √x√y

Proof: We have to find which statement must be true?

Given 0<y<x,

Option (a) :

Left hand side:

Left hand side is not equal to right hand side 

The statement is wrong. 

Option (b) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong.

Option (c) : 

Left hand side:

Right Hand side:

Left hand side is not equal to right hand side 

The statement is wrong.

Option (d) :

Left hand side:

Right Hand side:

Left hand side is equal to right hand side 

The statement is true.

Hence the correct choice is

Q.38. If 10^x = 64, what is the value of 10^x/2+1 ?

(a) 18

(b) 42

(c) 80

(d) 81

Proof: We have to find the value ofprovided

So,

By substituting we get

Hence the correct choice is.

Q.39.  is equal to

(a) 5/3

(b) −5/3

(c) 3/5

(d) −3/5

Proof: We have to simplify

Taking 5ⁿ as a common factor we get

Hence the correct alternative is.

Q.40. Ifthen=

(a) 3

(b) 9

(c) 27

(d) 81

Proof: We have to find

Given

Equating powers of rational exponents we get 

Substituting inwe get

Hence the correct choice is.

Fill in the Blanks Types of Questions(FBQs)

Q.1. (21² – 15²)^2/3 is equal to ________.

Proof:

= (6)²

= 36

Hence, (21² – 15²)^2/3 is equal to 36.

Q.2. 81^1/4 × 9^3/2 × 27^–4/3 is equal to _________.

Proof:

= 1

Hence, 81^1/4 × 9^3/2 × 27^–4/3 is equal to 1.

Q.3. = __________. 

Proof:

Hence,

Q.4. If x = 8^2/3 × 32^–2/5, then x^–5 = ________.

Proof: Let x =

⇒x =

⇒x =

⇒x =

⇒x =

⇒x = 1

Now,

= 1

Hence, if x = 8^2/3 × 32^–2/5, then x–5 = 1.

Q.5. If 6ⁿ = 1296, then 6^n–3 = _________.

Proof: Let 6ⁿ=1296

⇒6ⁿ = 64

⇒n = 4

Now,

6ⁿ⁻³=6⁴⁻³       

=6¹       

=6

Hence, if 6ⁿ = 1296, then 6^n–3 = 6.

Q.6. The value of 4 × (256)^–1/4 ÷ (243)^1/5 is ________.

Proof: Let x=

⇒x =

⇒x =

⇒x =

⇒x =

⇒x = 1 ÷ 3

⇒x = 1/3

Hence, the value of 4 × (256)–1/4 ÷ (243)1/5 is

Q.7. If (6x)⁶=6²³, then x = ________.

Proof: 

Hence, if

Q.8. = ___________.

Proof: 

(using the identity: a²−b²=(a+b)(a−b))

Hence,

Q.9. then x =_________.

Proof: 

⇒2p = x

⇒x = 2p

Hence, if then x=2p.

Q.10. If 5ⁿ⁺² = 625, then (12n + 3)^1/3 = _________.

Proof: Let 5ⁿ⁺²=625

⇒5ⁿ⁺² = 54

⇒n+2 =4

⇒n = 2

Now,

= 3

Hence, if 5ⁿ⁺² = 625, then (12n + 3)^1/3 = 3.

Q.11. If= __________.

Proof: 

=(a)^−a

Hence,

Q.12. If=3³ , then 5x + 6y = __________.

Proof: Given: =3³

⇒3^−5x = 3³

⇒−5x = 3

⇒5x = −3           ...(1)

(729)^y = 3³

⇒(36)^y = 3³

⇒36^y = 3³

⇒6y = 3               ...(2)

Adding (1) and (2), we get

5x + 6y=−3 + 3          

=0

Hence, 5x + 6y = 0.

Q.13. If 6^x–y = 36 and 3^x+y = 729, then x² – y² = _________.

Proof: Given: 6^x−y = 36

and 3^x+y = 729 

6^x−y = 36

⇒ 6^x−y = 6²

⇒x−y = 2              ...(1)

3^x+y = 729

⇒3^x+y = 3⁶

⇒x+y = 6               ...(2)

Adding (1) and (2), we get

2x = 8

⇒x = 4

Substituting the value of x in (2), we get

4+y = 6

⇒y = 2

Now,

x²−y² = 4²−2²          

=16 − 4          

=12

Hence, x² – y² = 12.

Q.14. equals __________.

Proof: 

=(2)^1/6

Hence, equals

Q.15. The productis equal to ________.

Proof: 

= (2)¹                     

= 2

Hence, the product is equal to 2.

Q.16. is equal to _________.

Proof: 

= (3)⁻²             

= 1/3²             

= 1/9

Hence,  is equal to

Q.17. The value of (256)^0.16  × (256)^0.09 is ________.

Proof: (256)^0.16 × (256)^0.09 = (256)^0.16+0.09

= (256)^0.25

= (2)²                            

= 4

Hence, the value of (256)^0.16  × (256)^0.09 is 4.

Very Short Answer Type Questions(VSAQs)

Q.1. Write (625)^−1/4 in decimal form.

Proof: We have to write (625)^−1/4 in decimal form. So,

Hence the decimal form ofis

Q.2. State the product law of exponents.

Proof: 

State the product law of exponents.

If is any real number and m ,n  are positive integers, then

By definition, we have

a^m x aⁿ = a^m+n

Thus the exponent "product rule" tells us that, when multiplying two powers that have the same base, we can add the exponents. 

Q.3. State the quotient law of exponents.

Proof: The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. If a is a non-zero real number and m, n are positive integers, then

We shall divide the proof into three parts 

(i) when m > n

(ii) when m = n

(iii) when m < n

Case 1

when m > n

We have 

Case 2

when m = n

We get

Cancelling common factors in numerator and denominator we get,

By definition we can write 1 as a⁰

Case 3

when m < n

In this case, we have 

Hence, whether m > n,m = n or, m < n

Q.4. State the power law of exponents.

Proof: The "power rule" tell us that to raise a power to a power, just multiply the exponents. 

If a is any real number and m, n are positive integers, then

We have,

factors

factors

Hence,

Q.5. If 2⁴ × 4² =16x, then find the value of x.

Proof: We have to find the value of x provided 2⁴ × 4² =16x

So,

By equating the exponents we get

Hence the value of x is .

Q.6. If 3^x-1 = 9 and 4^y+2 = 64, what is the value of x/y ?

Proof: We have to find the value offor

So,

By equating the exponent we get

Let’s take

By equating the exponent we get

By substituting x = 3, y = 1 we get 3 / 1

Hence the value of x/y is 3.

Q.7. Write the value of

Proof: We have to find the value of   So,

By using law rational exponentswe get

Hence the value ofis

Q.8. Writeas a rational number.

Proof: We have to find the value of

So,

Hence the value of the value ofis.

Q.9. Write the value of

Proof: We have to find the value of So,

= = 5×3 =15

Hence the value of the value of is

Q.10. For any positive real number x, find the value of

Proof: We have to find the value of L =

By using rational exponentswe get

By using rational exponentswe get 

By definition we can write x⁰ as 1

Hence the value of expression is.

Q.11. Write the value of

Proof: We have to find the value of So,

By using rational exponents  we get

Hence the simplified value ofis.

Q.12. Simplify

Proof: We have to simplify So,

Hence, the value ofis .

Q.13. For any positive real number x, write the value of

Proof: We have to simplifySo,

By using rational exponents, we get

Hence the value ofis x³.

Q.14. If (x − 1)³ = 8, What is the value of (x + 1)² ?

Proof: We have to find the value of  (x + 1)², where (x − 1)³ = 8

Consider (x − 1)³ =2³

By equating the base, we get

x - 2 = 1

x = 2 + 1

x = 3

By substituting x = 3 in (x + 1)²

^{= (x + 1)2}

= (3 + 1)^2is 

= 4²

= 16

Hence the value of(x + 1)² is 16.