Exponents And Powers
Class 8th Mathematics CBSE Solution
- Evaluate. (i) 3^-2 (ii) (-4)^-2 (iii) (1/2)^-5
- Simplify and express the result in power notation with positive exponent. (i)…
- Find the value of.(i) (3^circle + 4^-1) x 2^2 (ii) (2^-1 x 4^-1) / 2^-2 (iii)…
- Evaluate(i) 8^-1 x 5^3/2^-4 (ii) (5^-1 x 2^-1) x 6^-1
- Find the value of mfor which5m 5-3 = 5^5 .
- Evaluate (i) (ii)
- Simplify (i) (ii)
- Express the following numbers in standard form.(i) 0.0000000000085(ii)…
- Express the following numbers in usual form. (i) 3.02 x 10^-6 (ii) 4.5 x 10^4…
- Express the number appearing in the following statements in standard form.(i) 1…
- In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of…
Exercises 12.1
Question 1.Evaluate.
(i) 
(ii) 
(iii) 
Answer:
(i) Formula:
By using above formula in question:
(ii) Formula:
By using above formula in question:
(iii) Formula:
By using above formula in question:
Question 2.
Simplify and express the result in power notation with positive exponent.
(i) (-4)5 ÷ (-4)8
(ii) 
(iii) 
(iv) (3-7 ÷ 3-10) × 3-5
(v) 2-3 × (-7)-3
Answer:
(i) Formula: am ÷ an = a(m-n)
By using above formula in question:
(-4)5 ÷ (-4)8
= (-4)5-8
= (-4)-3
= 1/(-4)3
= -1/64
(ii) Formula: (am)n = amn
By using above formula in question:
(iii) Formula:- am × bm = (a × b)m
Using this formula:
(iv) Formula:- am ÷ an = a(m-n)
am × an = a{m+n}
By using above formula in question:
(3-7÷ 3-10) × 3-5
= (3 (-7+10)) × 3-5
= 33 × 3-5
= 3 (3-5)
= 3-2
= 1/32
= 1/9
(v) 2-3 × (-7)-3
By using the formula,
am × bm = (a × b)m
Question 3.
Find the value of.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Answer:
(i) Formula:
A0 = 1
By using above formula in question:
(30 + 4-1) × 22 = (1 + 1/4) × 22
⇒ (30 + 4-1) × 22 = (5/4) × 4
⇒ (30 + 4-1) × 22 = 5
(ii) Formula:
By using above formula in question:
(2-1 × 4-1) ÷ 2-2 = (2-1 × 2-2) ÷ 2-2
⇒ (2-1 × 4-1) ÷ 2-2 = (2-1 -2) ÷ 2-2
⇒ (2-1 × 4-1) ÷ 2-2 = (2-3) ÷ 2-2
⇒ (2-1 × 4-1) ÷ 2-2 = 2-3+2
⇒ (2-1 × 4-1) ÷ 2-2 = 2-1= 1/2
(iii) Formula:
By using above formula in question:
(iv) Formula:
By using above formula in question:
Since anything raise to the power 0 = 1
(v) Formula:
By using above formula in question:
Question 4.
Evaluate
(i) 
(ii) 
Answer:
(i) Formula
am/an = a(m-n)
By using above formula in question:
= 53 × 2
= 125 × 2
= 250
(ii) Formula:
By using above formula in question:
Question 5.
Find the value of m for which 5m ÷ 5-3 = 55.
Answer:
Formula:
= a(m-n)
am × an = a(m+n)
an = am, then n = m
By using above formula in question:
5m ÷ 5-3 = 55
⇒ 5m - (-3) = 55
⇒ 5m+3 = 55
⇒ m + 3 = 5
⇒ m = 2
Question 6.
Evaluate
(i) 
(ii) 
Answer:
(i) Formula:
By using above formula in question:
(ii) Formula:
By using above formula in question:
Question 7.
Simplify
(i) 
(ii) 
Answer:
(i) Formula:
By using above formula in question:
(ii) Formula:
By using above formula in question:
= 55
= 3125
Exercises 12.2
Question 1.Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
Answer:
(i) 0.0000000000085= 85 ×10-13 = 8.5 ×10-12
(ii) 0.00000000000942 = 942 ×10-14 = 9.42 ×10-12
(iii) 6020000000000000 = 602 ×1013 = 6.02 ×1015
(iv) 0.00000000837 = 837 × 10-11 = 8.37 × 10-9
(v) 31860000000 = 3186 × 107= 3.186 ×1010
Question 2.
Express the following numbers in usual form.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Answer:
We know by simple indices that:
am = a x a x a x a..........................m times
So, 25 = 2 x 2 x 2 x 2 x 2
And also that 
(i) 
(ii) 
(iii) 3 x 10-8 = 3/100000000 = .00000003
(iv) 1.001 x 109 = 1.0001 × 1000000000 = 1000100000
(v) 5.8 x 1012 = 5.8 × 1000000000000 = 5800000000000
(vi) 3.61492 x 106 = 3.61492 × 1000000 = 3614920
Question 3.
Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
Answer:
(i) 1 Micron = 10-6 m
(ii) 0.000,000,000,000,000,000,16 Coulomb
= 16/100000000000000000
= 16 × 10-18 = 1.6 × 10-19 Coulomb
(iii) 0.0000005 m
= 5/10000000
= 5 ×10-7 m
(iv)0.00001275 m
= 1275 /100000000
= 1275 × 10-8
=1.275 × 10-5
(v) 0.07 mm
= 7/100 mm
= 7 ×10-2 mm
Question 4.
In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.
Answer:
Thickness of 1 book = 20 mm
Thickness of 5 books = 5 x 20 mm = 100 mm
Thickness of 1 page = 0.016 mm
Thickness of 5 pages = 5 x 0.016 = 0.08 mm
Total Thickness = Thickness of 5 books + Thickness of 5 pages
Total Thickness = 100 + 0.08
Total Thickness = 100.08 mm