The document RD Sharma Solutions Ex-2.1, Exponents Of Real Numbers, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q.1. Simplify the following:**

**(i) 3(a ^{4} b^{3})^{10} x 5 (a^{2} b^{2})^{3}**

**Solution:** = 3(a^{40} b^{30}) x 5 (a^{6} b^{6}) = 15 (a^{46} b^{36})

**(ii) (2x ^{-2} y^{3})^{3}**

**Solution: ** = (2^{3} x ^{-2x3} y^{3x3}) = 8x ^{-6} y^{9}

**(iii) **

**Solution: **

**(iv) **

**Solution:** =

**(v)**

**Solution: **

**(vi) **

**Solution: ** = a^{16n−50}

**2. If a = 3 and b = -2, find the values of:**

**(i) a ^{a}+b^{b} **

**Solution:** (i) We have,a^{a}+b^{b} = 3^{3}+(−2)^{−2} = 3^{3}+ = 27+1/4 = 109/4

(ii) a^{b}+b^{a} = 3^{−2}+(−2)^{3} = (1/3)^{2}+(−2)^{3}

(iii) We have, a^{b}+b^{a} = (3+(−2))^{3(−2)} = (3−2))^{−6} = 1^{−6} = 1

**3.Prove that:** **(i) **

**(ii) (iii) **

**Solution: ** = 1

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Or, Therefore, LHS = RHS Hence proved

(ii) To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved (iii) To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

**4. Prove that:** **(i) **

**(ii)**

**Solution:** (i)

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

**5.Prove that:** **(i) **

**(ii) **

**Solution:** (i) To prove, = abc

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

abc

Therefore, LHS = RHS Hence proved (ii) To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

Therefore, LHS = RHS Hence proved

**6. If abc = 1, show that **

**Solution:** To prove,

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

We know abc = 1

By substituting the value c in equation (1), we get

Therefore, LHS = RHS Hence proved

**7. Simplify:** **(i) **

**Solution: **

**(ii)**

**Solution:** =

**(iii) **

**Solution:** =

**(iv) **

**Solution:**

**8. Solve the following equations for x: **

**(i) 7 ^{2x+3} = 1**

**Solution:** (i) We have,

⇒ 7^{2x+3} = 1 ⇒ 7^{2x+3} = 7^{0} ⇒ 2x + 3 = 0 ⇒ 2x = -3 ⇒ x

(ii) We have,

(iii) We have,

(iv) We have,

(v) We have,

**9. Solve the following equations for x:**

**(i) 2 ^{2x}−2^{x+3}+2^{4} = 0**

**(ii) 3 ^{2x+4}+1=2×3^{x+2}**

**Solution:**

(i) We have, ⇒ 2^{2x}−2^{x+3}+2^{4} = 0

⇒ 2^{2x}+2^{4}=2^{x}.2^{3}

⇒ Let ^{2x}=y

⇒ y^{2}+2^{4}=y×2^{3}

⇒ y^{2}−8y+16=0

⇒ y^{2}−4y−4y+16=0

⇒ y(y-4) -4(y-4) = 0

⇒ y = 4

⇒ x^{2}=2^{2}

⇒ x = 2

(ii) We have,

3^{2x+4}+1=2×3^{x+2}

(3^{x+2})^{2}+1=2×3^{x+2}

Let 3^{x+2} = y

y^{2}+1=2y

y^{2}−2y+1=0

y^{2}−y−y+1=0

y(y−1)−1(y−1)=0

(y−1)(y−1)=0

y = 1

**10. If 49392 = a ^{4}b^{2}c^{3}, find the values of a, b and c, where a, b and c, where a, b, and c are different positive primes.**

**Solution:** Taking out the LCM , the factors are 2^{4},3^{2 }and 7^{3} a^{4}b^{2}c^{3} = 2^{4},3^{2 }and 7^{3}

a = 2, b = 3 and c = 7 [Since, a, b and c are primes]

**11. If 1176 = 2 ^{a}×3^{b}×7^{c}, Find a, b, and c.**

**Solution:** Given that 2, 3 and 7 are factors of 1176. Taking out the LCM of 1176, we get 2^{3}×3^{1}×7^{2} = 2^{a}×3^{b}×7^{c} By comparing, we get

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

**12. Given 4725 = 3 ^{a}×5^{b}×7^{c}, find**

**Solution: **Taking out the LCM of 4725, we get

3^{3}×5^{2}×7^{1} = 3^{a}×5^{b}×7^{c}

By comparing, we get

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

**(ii) The value of 2 ^{−a}×3^{b}×7^{c}**

**Solution:**

= 2^{−3}×3^{2}×7^{1}

2^{−3}×3^{2}×7^{1} = 1/8×9×7

63/8

**13. If a = xy ^{p−1}, b = xy^{q−1} and c = xy^{r−1}, prove that a^{q−r}b^{r−p}c^{p−q} = 1**

**Solution:** Given, a = xy^{p−1}, b = xy^{q−1} and c = xy^{r−1}

To prove, a^{q−r}b^{r−p}c^{p−q} = 1

Left hand side (LHS) = Right hand side (RHS) Considering LHS, = a^{q−r}b^{r−p}c^{p−q} ……(i)

By substituting the value of a, b and c in equation

(i), we get

= (xy^{p−1})^{q−r}(xy^{q−1})r−p(xy^{r−1})^{p−q}

= xy^{pq−pr−q+r}xy^{qr−pq−r+p}xy^{rp−rq−p+q}

= xy^{pq−pr−q+r+qr−pq−r+p+rp−rq−p+q}

= xy^{0} = 1 * *

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