**Frank Learning Maths Class 8 Chapter 2 Exponents Chapter Check-up Solutions**

This post is created to help all the CBSE Class 8 students for the Solutions of Frank Learning Maths Class 8 Mathematics Book, Chapter 2 Exponents Chapter Check-up. Here students can easily find step by step solutions of all the problems for Exponents Chapter Check-up. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily.

**Chapter 2 – Exponents**

**Exponents – Chapter Check-up all Questions Solution**

**(1) Fill in the blanks**

**Solution : **

**(a)** The reciprocal of (2/5)^-1 is – (2/5)

**(b)** (- 2)^2×3-1 = (- 2)^6-1 = (- 2)^5 = – 3^2

**(c)** (- 3/4)^5 × (5/3)^5 is equal to – 3^5/4^5 × 5^5/3^5 = (- 5/4)^5

**(d)** [2^-1 + 3^-1 + 4^-1]^0 = 1

**(e)** (6^0 – 7^0) (6^0 + 7^0) = (1 – 1) (1 + 1) = 0

**(f)** The standard form of 0.000064 is 6.4 × 10^-5

**(g)** The standard form of 234500000 is 2.345 × 10^8

**(h)** The usual form of 2.03 × 10^-5 is = 0.000203

**(i)** Cube of – 1/2 is (- 1/2)^3 = – 1/8

**(j)** The value of [3^-1 × 4^-1]^2 is 1/3 × 1/4 = 1/12

**(2) Simplify**

**Solution : **

**(a)** (1/4)^{-2} + (1/2)^{-2} + (1/3) ^{-2}

= (4) ^{2} + (2) ^{2} + (3)^{2}

= 16 + 4 + 9

= 29…(Simplified)

**(b)** (-2/3) ^{-2})^{3} × (1/3)^{-4} × 3^{-1} × 1/6

= ((3/2) ^{2}) ^{3} × (3) ^{4} × 1/3 × 1/3×2

= (3 × 2/2 × 2) ^{3} × 3^{2} × 3^{2} × 1/3^{2} × 2

= 3^{3} × 3^{3}/2^{3} × 2^{3} × 3^{2}/2

= 3^{8}/2^{8}…(Simplified)

**(c)** 49 × z^{-3}/7^{-3} × 10 × z^{-5} (z ≠ 0)

= 7^{2} × z^{2}/7^{-3} × 10

= 7^{5} × z^{2} × 10^{-1}…(Simplified)

**(d)** (2^{5} ÷ 2^{8}) × 2^{-7}

= 2^{5}/2^{8} × 2^{-7}

= 2^{-7}/2^{-3}

= 1/2^{4}…(Simplified)

**(3) Find the value of x so that**

**Solution : **

**(a)** (5/3) ^{-2} × (5/3) ^{-14 }= (5/3) ^{8x}^{
}

= (5/3) ^{-16 }= (5/3) ^{8x
}

= 8x = -16

= x = -2

So, the value of x will be -2

**(b)** (- 2) ^{3} × (- 2) ^{-6} = (- 2) ^{2x – 1}^{
}

= (- 2) ^{-3} = (- 2) ^{2x – 1
}

= 2x -1 = -3

= 2x = -2

= x = -1

Hence, the value of x will be -1

**(c)** (2^{-1} + 4^{-1} + 6^{-1} + 8^{-1}) ^{x} = 1

= (2^{-1} + 4^{-1} + 6^{-1} + 8^{-1}) ^{x}

= (2^{-1} + 4^{-1} + 6^{-1} + 8^{-1}) ^{0
}

= x = 0

Therefore, the value of x will be 0.

**(4) Express each of the following in standard form**

**Solution : **

**(a)** The mass of a proton in gram is 1673/1000000000000000000000000000

= 1.673 × 10^{-24 }gram

**(b)** A Helium atom has a diameter of 0.000000022 cm.

= 2.2 × 10^{-8} cm

**(c)** Mass of a molecule of hydrogen gas is about 0.00000000000000000000334 tons.

= 3.34 × 10^{-22} tons.

**(d)** Human body has 1 trillion of cells which vary in shapes and sizes.

= 1 × 10^{12}

**(e)** Express 56 km in m.

= 5.6 × 10^{4} m

**(f)** Express 5 tons in g.

= 5 × 10^{5} gm

**(g)** Express 2 years in seconds.

= 6.3072 × 10^{7} s

**(h)** Express 5 hectares in cm^{2} (1 hectare = 10000 m^{2})

= 5 × 10^{8} cm^{2}

**(5) Find the value of n**

**Solution : **

**(a)** 6^{n}/6^{-2} = 6^{3}

= (6) ^{n+2 }= 6^{3}

= x + 2 = 3

= x = 1

Hence, the value of n will be 1.

**(b)** 2^{n }× 2^{6}/2^{-3} = 2^{18}

= 2^{6+n+3 }= 2^{18}

= n + 9 = 18

= n = 9

Thus, the value of n will be 9.

**Question no – (6) **

**Solution : **

As per the given question,

The diameter of the sun = 1.4 × 10^{9} m

The diameter of the earth = 1.2756 × 10^{7} m

**∴** 1.4 × 10^{9} m

= 1.2756 × 10^{7} m

= 1.3 × 10^{7} m (Approximately)

Therefore, diameter of the sun is 100 time of the Earth diameter.

**Question no – (7) **

**Solution : **

According to the question,

Mass of mars is 6.42 × 10^{29} kg

Mass of the sun is 1.99 × 10^{30} kg.

**∴** 6.42 × 10^{29} kg + 1.99 × 10^{30}

= 6.42 × 10^{29} + 19.9 × 10^{29}

= 26.32 × 10^{29}

= 2.632 × 10^{30} kg

Hence, the total mass will be 2.632 × 10^{30} kg.

**Question no – (8) **

**Solution : **

As per the question,

Distance between sun and earth is = 1.496 × 10^{8} km

Distance between earth and moon is = 3.84 × 10^{8} m

∴ Distance between moon and sun,

= – 1.496 × 10^{8}/2

= 7.48 × 10^{7}

Therefore, the distance between moon and sun will be 7.48 × 10^{7}

**Question no – (11) **

**Solution : **

According to the question,

The cells of a bacteria double itself every hour.

How many cells will there be after 8 hours,

if initially we start with 1 cell.

The answer in power will be__ 2 ^{8}__

**Question no – (12) **

**Solution : **

Given, (2/9)^{3} × (2/9)^{-6} = (2/9)^{2x-1}

= – 3 = 2x – 1

= 2x = -2

= x = -1

Therefore, the x will be -1

**Question no – (13) **

**Solution : **

Let, we should divided the number by x

∴ (- 3/2) ^{-3}/x = (4/27) ^{-2
}

= -(2/3) ^{3} × 1/x = (27/4) ^{2
}

= – (2/3) ^{3} × (4/27) ^{2} = x

= – 2^{3}/3^{3} × 2^{4}/3^{6} = x

= (- 2) ^{7}/39 = x

Therefore, (-2)^{7}/39 should be divided.