The document RD Sharma Solutions Ex-2.2, (Part - 2), 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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show that x^{3} − 6x = 6

x^{3 }− 6x = 6

Putting cube on both the sides, we get

As we know, (a+b)3 = a^{3} + b^{3} + 3ab(a+b)

x^{3} = 6 + 6x

x^{3} – 6x = 6

Hence proved

**12. Determine (8x) ^{x}, if 9^{x+2 }= 240 + 9^{x}.**

9^{x+2 }= 240 + 9^{x}

9^{x} .9^{2} = 240 + 9^{x}

Let 9^{x} be y

81y = 240 + y

81y – y = 240

80y = 240

y = 3

Since, y = 3

Then,

9^{x} = 3

3^{2x} = 3

Therefore, x = 1/2

= 2

Therefore (8x)^{x} = 2

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

3^{x+1} = 9^{x-2}

3^{x+1} = 3^{2x-4}

x + 1 = 2x – 4

x = 5

Therefore the value of 2^{1+x} = 2^{1+5} = 2^{6} = 64

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

3^{4x} = (81)^{-1} and (10)^{1/y} = 0.0001

3^{4x} = (3)^{-4}

x = -1

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

(10)^{1/y} = (10)^{−4}

To find the value of 2^{-x+4y}, we need to substitute the value of x and y

**15. If 5 ^{3x} = 125 and 10^{y} = 0.001. Find x and y.**

5^{3x} = 125 and 10^{y} = 0.001

5^{3x} = 5^{3}

x = 1

Now,

10^{y} = 0.001

10^{y} = 10^{-3}

y = -3

Therefore, the value of x = 1 and the value of y = -3

**16. Solve the following equations**

(i) 3^{x+1} = 2^{7×34}

3^{x+1} = 3^{3}×3^{4}

3^{x+1} = 3^{3+4}

x + 1 = 3 + 4 [By equating exponents]

x + 1 = 7

x = 7 − 1

x = 6

2^{4x} = 2^{3}

4x = 3 (By equating exponents)

x = 3/4

(By equating exponents)

(iii). 3^{x−1}×5^{2y−3 }= 225

3^{x−1}×5^{2y−3 }= 3^{2}×5^{2}

x−1 = 2 [By equating exponents]

x = 3

3^{x−1}×5^{2y−3} = 3^{2}×5^{2}

2y−3 = 2 [By equating exponents]

2y = 5

y = 5/2

3x+3 = 4y+8 —eq1

3(3x+3) = 6+2x+24

9x+9 = 30+2x

7x = 21

x = 21/7

x = 3

Putting value of x in eq2

2x−2−3+2x = −3x [By equating exponents]

4x+3x = 5

7x = 5

x = 5/7

[By equating exponents]

**17. If a and b are distinct positive primes such that *** ***find x and y**

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

a^{x}b^{y} find x and y

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

=(ab)^{−1 }= a^{−1}b^{−1 }

By equating exponents

x = −1,y = −1

Therefore x+y+2 = −1−1+2 = 0

**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}**

**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**

We have to find the value of

= 24/49

**21. Simplify**

**22. Show that**

Hence LHS = RHS

**23. **

(i)

= x^{0} y^{0}

= 1

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