Question

2p-1
Q4.Find the value of p for which
(2 upon7)^5 ×(2upon7)^-6= (2upon7) ^2p-1​

Answer 1

SoluTion :-

$\sf {\frac{2}{7}^{5} \times \frac{2}{7}^{-6} = \frac{2}{7}^{2p-1}}$

Since the bases are same, the powers will be added (multiplication):-

$\sf {\frac{2}{7}^{(5-6)}=\frac{2}{7}^{2p-1}}$

Since the bases are same, the powers will be equal :-

5 - 6 = 2p - 1

-1 = 2p - 1

-1 + 1 = 2p

0 = 2p

0 = p

Answer - Value of p = 0

Answer 2

Answer:

Exercise 2.1

Solve the following equations:

Question 1:

x - 2 = 7

Solution:

Given, x – 2 = 7

By adding 2 to both sides, we get

x – 2 + 2 = 7 + 2

⇒ x = 7 + 2

⇒ x = 9

Question 2:

y + 3 = 10

Solution:

Given,y + 3 = 10

By subtracting 3 from both sides, we get

y + 3 – 3 = 10 – 3

⇒ y = 10 – 3

⇒ y = 7

Question 3:

6 = z + 2

Solution:

Given,6 = z + 2

By substracting 2 from both sides, we get

6 – 2 = z + 2 – 2

⇒4 = z + 2 – 2

⇒ 4 = z

⇒ z = 4

Question 4:

Solution:

Given,

By subtracting from both sides, we get

Question 5:

6x = 12

Solution:

Given,6x = 12

After dividing both sides by 6, we get

Question 6:

Solution:

Given,

After multiplying both sides by 5, we get

Question 7:

Solution:

Given,

By multiplying both sides with 3, we get

After dividing both sides by 2, we get

Question 8:

Solution:

Given,

After multiplying both sides by 1.5, we get

Question 9:

7x – 9 = 16

Solution:

Given,7x – 9 = 16

By adding 9 to both sides, we get

After dividing both sides by 7, we get

Question 10:

Solution:

Given,14y – 18 = 13

By adding 8 to both sides, we get

14y – 8 + 8 = 13 + 8

⇒ 14y = 21

After dividing both sides by 14, we get

Question 11:

17 + 6p = 9

Solution:

Given, 17 + 6p = 9

After subtracting 17 from both sides, we get

17 + 6p – 17 = 9 – 17

⇒ 6p = –8

After dividing both sides by 6, we get

Question 12:

Solution:

Given,

By transposing 1 from LHS to RHS, we get

After multiplying both sides with 3, we get

Exercise 2.2

Question 1:

If you subtract from a number and multiply the result by , you get . What is the number?

Solution:

Let the number is m.

As per question we have

After dividing both sides by we get

By transposing to RHS, we get

Question 2:

The perimeter of a rectangular swimming pool is 154m. Its length is 2m more than twice its breadth.

What are the length and the breadth of the pool?

Solution:

Given, perimeter of the rectangular swimming pool = 154m

Length is 2 metre more than the breadth.

Let the breadth of the swimming pool = a metre

Therefore, as per question, length of the swimming pool = (2a + 2) metre

We know that,

perimeter of rectangle = 2 (length + breadth)

Therefore, 154m = 2[(2a + 2) + a]

⇒ 154m = 2(2a + 2 + a )

⇒ 154m = 2 (3a + 2)

⇒ 154m = 6a + 4

By subtracting 4 from both sides, we get

154m – 4 = 6a + 4 – 4

⇒ 150m = 6a

After dividing both sides by 6, we get