{"id":546008,"date":"2021-10-05T11:12:00","date_gmt":"2021-10-05T11:12:00","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546008"},"modified":"2021-10-07T04:42:19","modified_gmt":"2021-10-07T04:42:19","slug":"rd-sharma-solutions-for-class-8-maths-chapter-2-powers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-2-powers\/","title":{"rendered":"RD Sharma Solutions for Class 8 Maths Chapter 2\u2013Powers"},"content":{"rendered":"\n

Class 8: Maths Chapter 2 solutions. Complete Class 8 Maths Chapter 2 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 2\u2013Powers<\/h2>\n\n\n\n

RD Sharma 8th Maths Chapter 2, Class 8 Maths Chapter 2 solutions<\/p>\n\n\n\n

EXERCISE 2.1 PAGE NO: 2.8<\/h4>\n\n\n\n

1. Express each of the following as a rational number of the form p\/q, where p and q are integers and q \u2260 0:<\/strong><\/p>\n\n\n\n

(i) 2-3<\/sup><\/strong><\/p>\n\n\n\n

(ii) (-4)-2<\/sup><\/strong><\/p>\n\n\n\n

(iii) 1\/(3)-2<\/sup><\/strong><\/p>\n\n\n\n

(iv) (1\/2)-5<\/sup><\/strong><\/p>\n\n\n\n

(v) (2\/3)-2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> 2-3<\/sup> = 1\/23<\/sup> = 1\/2\u00d72\u00d72 = 1\/8 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> (-4)-2<\/sup> = 1\/-42<\/sup> = 1\/-4\u00d7-4 = 1\/16 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(iii)<\/strong> 1\/(3)-2<\/sup> = 32<\/sup> = 3\u00d73 = 9 (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> (1\/2)-5<\/sup> = 25 <\/sup>\/ 15<\/sup> = 2\u00d72\u00d72\u00d72\u00d72 = 32 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(v)<\/strong> (2\/3)-2<\/sup> = 32<\/sup> \/ 22<\/sup> = 3\u00d73 \/ 2\u00d72 = 9\/4 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

2. Find the values of each of the following:<\/strong><\/p>\n\n\n\n

(i) 3-1<\/sup> + 4-1<\/sup><\/strong><\/p>\n\n\n\n

(ii) (30<\/sup> + 4-1<\/sup>) \u00d7 22<\/sup><\/strong><\/p>\n\n\n\n

(iii) (3-1<\/sup> + 4-1<\/sup> + 5-1<\/sup>)0<\/sup><\/strong><\/p>\n\n\n\n

(iv) ((1\/3)-1<\/sup> \u2013 (1\/4)-1<\/sup>)-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> 3-1<\/sup> + 4-1<\/sup><\/p>\n\n\n\n

1\/3 + 1\/4 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

LCM of 3 and 4 is 12<\/p>\n\n\n\n

(1\u00d74 + 1\u00d73)\/12<\/p>\n\n\n\n

(4+3)\/12<\/p>\n\n\n\n

7\/12<\/p>\n\n\n\n

(ii)<\/strong> (30<\/sup> + 4-1<\/sup>) \u00d7 22<\/sup><\/p>\n\n\n\n

(1 + 1\/4) \u00d7 4 (we know that a-n<\/sup> = 1\/an<\/sup>, a0<\/sup> = 1)<\/p>\n\n\n\n

LCM of 1 and 4 is 4<\/p>\n\n\n\n

(1\u00d74 + 1\u00d71)\/4 \u00d7 4<\/p>\n\n\n\n

(4+1)\/4 \u00d7 4<\/p>\n\n\n\n

5\/4 \u00d7 4<\/p>\n\n\n\n

5<\/p>\n\n\n\n

(iii) <\/strong>(3-1<\/sup> + 4-1<\/sup> + 5-1<\/sup>)0<\/sup><\/p>\n\n\n\n

(We know that a0<\/sup> = 1)<\/p>\n\n\n\n

(3-1<\/sup> + 4-1<\/sup> + 5-1<\/sup>)0<\/sup> = 1<\/p>\n\n\n\n

(iv)<\/strong> ((1\/3)-1<\/sup> \u2013 (1\/4)-1<\/sup>)-1<\/sup><\/p>\n\n\n\n

(31<\/sup> \u2013 41<\/sup>)-1<\/sup> (we know that 1\/a-n<\/sup> = an<\/sup>, a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(3-4)-1<\/sup><\/p>\n\n\n\n

(-1)-1<\/sup><\/p>\n\n\n\n

1\/-1 = -1<\/p>\n\n\n\n

3. Find the values of each of the following:<\/strong><\/p>\n\n\n\n

(i) (1\/2)-1<\/sup> + (1\/3)-1<\/sup> + (1\/4)-1<\/sup><\/strong><\/p>\n\n\n\n

(ii) (1\/2)-2<\/sup> + (1\/3)-2<\/sup> + (1\/4)-2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (2-1<\/sup> \u00d7 4-1<\/sup>) \u00f7 2-2<\/sup><\/strong><\/p>\n\n\n\n

(iv) (5-1<\/sup> \u00d7 2-1<\/sup>) \u00f7 6-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (1\/2)-1<\/sup> + (1\/3)-1<\/sup> + (1\/4)-1<\/sup><\/p>\n\n\n\n

21<\/sup> + 31<\/sup> + 41<\/sup> (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

2+3+4 = 9<\/p>\n\n\n\n

(ii)<\/strong> (1\/2)-2<\/sup> + (1\/3)-2<\/sup> + (1\/4)-2<\/sup><\/p>\n\n\n\n

22<\/sup> + 32<\/sup> + 42<\/sup> (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

2\u00d72 + 3\u00d73 + 4\u00d74<\/p>\n\n\n\n

4+9+16 = 29<\/p>\n\n\n\n

(iii)<\/strong> (2-1<\/sup> \u00d7 4-1<\/sup>) \u00f7 2-2<\/sup><\/p>\n\n\n\n

(1\/21<\/sup> \u00d7 1\/41<\/sup>) \/ (1\/22<\/sup>) (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/2 \u00d7 1\/4) \u00d7 4\/1 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

1\/2<\/p>\n\n\n\n

(iv)<\/strong> (5-1<\/sup> \u00d7 2-1<\/sup>) \u00f7 6-1<\/sup><\/p>\n\n\n\n

(1\/51<\/sup> \u00d7 1\/21<\/sup>) \/ (1\/61<\/sup>) (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/5 \u00d7 1\/2) \u00d7 6\/1 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

3\/5<\/p>\n\n\n\n

4. Simplify:<\/strong><\/p>\n\n\n\n

(i) (4-1<\/sup> \u00d7 3-1<\/sup>)2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (5-1<\/sup> \u00f7 6-1<\/sup>)3<\/sup><\/strong><\/p>\n\n\n\n

(iii) (2-1<\/sup> + 3-1<\/sup>)-1<\/sup><\/strong><\/p>\n\n\n\n

(iv) (3-1<\/sup> \u00d7 4-1<\/sup>)-1<\/sup> \u00d7 5-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (4-1<\/sup> \u00d7 3-1<\/sup>)2<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/4 \u00d7 1\/3)2<\/sup><\/p>\n\n\n\n

(1\/12)2<\/sup><\/p>\n\n\n\n

(1\u00d71 \/ 12\u00d712)<\/p>\n\n\n\n

1\/144<\/p>\n\n\n\n

(ii)<\/strong> (5-1<\/sup> \u00f7 6-1<\/sup>)3<\/sup><\/p>\n\n\n\n

((1\/5) \/ (1\/6))3<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

((1\/5) \u00d7 6)3<\/sup> (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

(6\/5)3<\/sup><\/p>\n\n\n\n

6\u00d76\u00d76 \/ 5\u00d75\u00d75<\/p>\n\n\n\n

216\/125<\/p>\n\n\n\n

(iii)<\/strong> (2-1<\/sup> + 3-1<\/sup>)-1<\/sup><\/p>\n\n\n\n

(1\/2 + 1\/3)-1<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

LCM of 2 and 3 is 6<\/p>\n\n\n\n

((1\u00d73 + 1\u00d72)\/6)-1<\/sup><\/p>\n\n\n\n

(5\/6)-1<\/sup><\/p>\n\n\n\n

6\/5<\/p>\n\n\n\n

(iv)<\/strong> (3-1<\/sup> \u00d7 4-1<\/sup>)-1<\/sup> \u00d7 5-1<\/sup><\/p>\n\n\n\n

(1\/3 \u00d7 1\/4)-1<\/sup> \u00d7 1\/5 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/12)-1<\/sup> \u00d7 1\/5<\/p>\n\n\n\n

12\/5<\/p>\n\n\n\n

5. Simplify:<\/strong><\/p>\n\n\n\n

(i) (32<\/sup> + 22<\/sup>) \u00d7 (1\/2)3<\/sup><\/strong><\/p>\n\n\n\n

(ii) (32<\/sup> \u2013 22<\/sup>) \u00d7 (2\/3)-3<\/sup><\/strong><\/p>\n\n\n\n

(iii) ((1\/3)-3<\/sup> \u2013 (1\/2)-3<\/sup>) \u00f7 (1\/4)-3<\/sup><\/strong><\/p>\n\n\n\n

(iv) (22<\/sup> + 32<\/sup> \u2013 42<\/sup>) \u00f7 (3\/2)2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (32<\/sup> + 22<\/sup>) \u00d7 (1\/2)3<\/sup><\/p>\n\n\n\n

(9 + 4) \u00d7 1\/8 = 13\/8<\/p>\n\n\n\n

(ii)<\/strong> (32<\/sup> \u2013 22<\/sup>) \u00d7 (2\/3)-3<\/sup><\/p>\n\n\n\n

(9-4) \u00d7 (3\/2)3<\/sup><\/p>\n\n\n\n

5 \u00d7 (27\/8)<\/p>\n\n\n\n

135\/8<\/p>\n\n\n\n

(iii)<\/strong> ((1\/3)-3<\/sup> \u2013 (1\/2)-3<\/sup>) \u00f7 (1\/4)-3<\/sup><\/p>\n\n\n\n

(33 <\/sup>\u2013 23<\/sup>) \u00f7 43<\/sup> (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

(27-8) \u00f7 64<\/p>\n\n\n\n

19 \u00d7 1\/64 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

19\/64<\/p>\n\n\n\n

(iv)<\/strong> (22<\/sup> + 32<\/sup> \u2013 42<\/sup>) \u00f7 (3\/2)2<\/sup><\/p>\n\n\n\n

(4 + 9 \u2013 16) \u00f7 (9\/4)<\/p>\n\n\n\n

(-3) \u00d7 4\/9 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

-4\/3<\/p>\n\n\n\n

6. By what number should 5-1<\/sup> be multiplied so that the product may be equal to (-7)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, 5-1<\/sup> \u00d7 x = (-7)-1<\/sup><\/p>\n\n\n\n

1\/5 \u00d7 x = 1\/-7<\/p>\n\n\n\n

x = (-1\/7) \/ (1\/5)<\/p>\n\n\n\n

= (-1\/7) \u00d7 (5\/1)<\/p>\n\n\n\n

= -5\/7<\/p>\n\n\n\n

7. By what number should (1\/2)-1<\/sup> be multiplied so that the product may be equal to (-4\/7)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, (1\/2)-1<\/sup> \u00d7 x = (-4\/7)-1<\/sup><\/p>\n\n\n\n

1\/(1\/2) \u00d7 x = 1\/(-4\/7)<\/p>\n\n\n\n

x = (-7\/4) \/ (2\/1)<\/p>\n\n\n\n

= (-7\/4) \u00d7 (1\/2)<\/p>\n\n\n\n

= -7\/8<\/p>\n\n\n\n

8. By what number should (-15)-1<\/sup> be divided so that the quotient may be equal to (-5)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, (-15)-1<\/sup> \u00f7 x = (-5)-1<\/sup><\/p>\n\n\n\n

1\/-15 \u00d7 1\/x = 1\/-5<\/p>\n\n\n\n

1\/x = (1\u00d7-15)\/-5<\/p>\n\n\n\n

1\/x = 3<\/p>\n\n\n\n

x = 1\/3<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 2.2 PAGE NO: 2.18<\/h4>\n\n\n\n

1. Write each of the following in exponential form:<\/strong><\/p>\n\n\n\n

(i) (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup><\/strong><\/p>\n\n\n\n

(ii) (2\/5)-2<\/sup> \u00d7 (2\/5)-2<\/sup> \u00d7 (2\/5)-2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup> \u00d7 (3\/2)-1<\/sup><\/p>\n\n\n\n

(3\/2)-4<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>, an<\/sup> = a\u00d7a\u2026n times)<\/p>\n\n\n\n

(ii)<\/strong> (2\/5)-2<\/sup> \u00d7 (2\/5)-2<\/sup> \u00d7 (2\/5)-2<\/sup><\/p>\n\n\n\n

(2\/5)-6<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>, an<\/sup> = a\u00d7a\u2026n times)<\/p>\n\n\n\n

2. Evaluate:<\/strong><\/p>\n\n\n\n

(i) 5-2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (-3)-2<\/sup><\/strong><\/p>\n\n\n\n

(iii) (1\/3)-4<\/sup><\/strong><\/p>\n\n\n\n

(iv) (-1\/2)-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> 5-2<\/sup><\/p>\n\n\n\n

1\/52<\/sup> = 1\/25 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> (-3)-2<\/sup><\/p>\n\n\n\n

(1\/-3)2<\/sup> = 1\/9 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(iii)<\/strong> (1\/3)-4<\/sup><\/p>\n\n\n\n

34<\/sup> = 81 (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> (-1\/2)-1<\/sup><\/p>\n\n\n\n

-21<\/sup> = -2 (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

3. Express each of the following as a rational number in the form p\/q:<\/strong><\/p>\n\n\n\n

(i) 6-1<\/sup><\/strong><\/p>\n\n\n\n

(ii) (-7)-1<\/sup><\/strong><\/p>\n\n\n\n

(iii) (1\/4)-1<\/sup><\/strong><\/p>\n\n\n\n

(iv) (-4)-1<\/sup> \u00d7 (-3\/2)-1<\/sup><\/strong><\/p>\n\n\n\n

(v) (3\/5)-1<\/sup> \u00d7 (5\/2)-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> 6-1<\/sup><\/p>\n\n\n\n

1\/61<\/sup> = 1\/6 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> (-7)-1<\/sup><\/p>\n\n\n\n

1\/-71<\/sup> = -1\/7 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(iii)<\/strong> (1\/4)-1<\/sup><\/p>\n\n\n\n

41<\/sup> = 4 (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> (-4)-1<\/sup> \u00d7 (-3\/2)-1<\/sup><\/p>\n\n\n\n

1\/-41<\/sup> \u00d7 (2\/-3)1<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>, 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

1\/-2 \u00d7 -1\/3<\/p>\n\n\n\n

1\/6<\/p>\n\n\n\n

(v)<\/strong> (3\/5)-1<\/sup> \u00d7 (5\/2)-1<\/sup><\/p>\n\n\n\n

(5\/3)1<\/sup> \u00d7 (2\/5)1<\/sup><\/p>\n\n\n\n

5\/3 \u00d7 2\/5<\/p>\n\n\n\n

2\/3<\/p>\n\n\n\n

4. Simplify:<\/strong><\/p>\n\n\n\n

(i) (4-1<\/sup> \u00d7 3-1<\/sup>)2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (5-1<\/sup> \u00f7 6-1<\/sup>)3<\/sup><\/strong><\/p>\n\n\n\n

(iii) (2-1<\/sup> + 3-1<\/sup>)-1<\/sup><\/strong><\/p>\n\n\n\n

(iv) (3-1<\/sup> \u00d7 4-1<\/sup>)-1<\/sup> \u00d7 5-1<\/sup><\/strong><\/p>\n\n\n\n

(v) (4-1<\/sup> \u2013 5-1<\/sup>) \u00f7 3-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (4-1<\/sup> \u00d7 3-1<\/sup>)2<\/sup><\/p>\n\n\n\n

(1\/4 \u00d7 1\/3)2<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/12)2<\/sup><\/p>\n\n\n\n

1\/144<\/p>\n\n\n\n

(ii)<\/strong> (5-1<\/sup> \u00f7 6-1<\/sup>)3<\/sup><\/p>\n\n\n\n

(1\/5 \u00f7 1\/6)3<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/5 \u00d7 6)3<\/sup> (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

(6\/5)3<\/sup><\/p>\n\n\n\n

216\/125<\/p>\n\n\n\n

(iii)<\/strong> (2-1<\/sup> + 3-1<\/sup>)-1<\/sup><\/p>\n\n\n\n

(1\/2 + 1\/3)-1<\/sup> (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

LCM of 2 and 3 is 6<\/p>\n\n\n\n

((3+2)\/6)-1<\/sup><\/p>\n\n\n\n

(5\/6)-1<\/sup> (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

6\/5<\/p>\n\n\n\n

(iv)<\/strong> (3-1<\/sup> \u00d7 4-1<\/sup>)-1<\/sup> \u00d7 5-1<\/sup><\/p>\n\n\n\n

(1\/3 \u00d7 1\/4)-1<\/sup> \u00d7 1\/5 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

(1\/12)-1<\/sup> \u00d7 1\/5 (we know that 1\/a-n<\/sup> = an<\/sup>)<\/p>\n\n\n\n

12 \u00d7 1\/5<\/p>\n\n\n\n

12\/5<\/p>\n\n\n\n

(v)<\/strong> (4-1<\/sup> \u2013 5-1<\/sup>) \u00f7 3-1<\/sup><\/p>\n\n\n\n

(1\/4 \u2013 1\/5) \u00f7 1\/3 (we know that a-n<\/sup> = 1\/an<\/sup>)<\/p>\n\n\n\n

LCM of 4 and 5 is 20<\/p>\n\n\n\n

(5-4)\/20 \u00d7 3\/1 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

1\/20 \u00d7 3<\/p>\n\n\n\n

3\/20<\/p>\n\n\n\n

5. Express each of the following rational numbers with a negative exponent:<\/strong><\/p>\n\n\n\n

(i) (1\/4)3<\/sup><\/strong><\/p>\n\n\n\n

(ii) 35<\/sup><\/strong><\/p>\n\n\n\n

(iii) (3\/5)4<\/sup><\/strong><\/p>\n\n\n\n

(iv) ((3\/2)4<\/sup>)-3<\/sup><\/strong><\/p>\n\n\n\n

(v) ((7\/3)4<\/sup>)-3<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (1\/4)3<\/sup><\/p>\n\n\n\n

(4)-3<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> 35<\/sup><\/p>\n\n\n\n

(1\/3)-5<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(iii)<\/strong> (3\/5)4<\/sup><\/p>\n\n\n\n

(5\/3)-4<\/sup> (we know that (a\/b)-n<\/sup> = (b\/a)n<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> ((3\/2)4<\/sup>)-3<\/sup><\/p>\n\n\n\n

(3\/2)-12<\/sup> (we know that (an<\/sup>)m<\/sup> = anm<\/sup>)<\/p>\n\n\n\n

(v)<\/strong> ((7\/3)4<\/sup>)-3<\/sup><\/p>\n\n\n\n

(7\/3)-12<\/sup> (we know that (an<\/sup>)m<\/sup> = anm<\/sup>)<\/p>\n\n\n\n

6. Express each of the following rational numbers with a positive exponent:<\/strong><\/p>\n\n\n\n

(i) (3\/4)-2<\/sup><\/strong><\/p>\n\n\n\n

(ii) (5\/4)-3<\/sup><\/strong><\/p>\n\n\n\n

(iii) 43<\/sup> \u00d7 4-9<\/sup><\/strong><\/p>\n\n\n\n

(iv) ((4\/3)-3<\/sup>)-4<\/sup><\/strong><\/p>\n\n\n\n

(v) ((3\/2)4<\/sup>)-2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> (3\/4)-2<\/sup><\/p>\n\n\n\n

(4\/3)2<\/sup> (we know that (a\/b)-n<\/sup> = (b\/a)n<\/sup>)<\/p>\n\n\n\n

(ii)<\/strong> (5\/4)-3<\/sup><\/p>\n\n\n\n

(4\/5)3<\/sup> (we know that (a\/b)-n<\/sup> = (b\/a)n<\/sup>)<\/p>\n\n\n\n

(iii)<\/strong> 43<\/sup> \u00d7 4-9<\/sup><\/p>\n\n\n\n

(4)3-9<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

4-6<\/sup><\/p>\n\n\n\n

(1\/4)6<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(iv)<\/strong> ((4\/3)-3<\/sup>)-4<\/sup><\/p>\n\n\n\n

(4\/3)12 <\/sup>(we know that (an<\/sup>)m<\/sup> = anm<\/sup>)<\/p>\n\n\n\n

(v)<\/strong> ((3\/2)4<\/sup>)-2<\/sup><\/p>\n\n\n\n

(3\/2)-8<\/sup> (we know that (an<\/sup>)m<\/sup> = anm<\/sup>)<\/p>\n\n\n\n

(2\/3)8<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

7. Simplify:<\/strong><\/p>\n\n\n\n

(i) ((1\/3)-3<\/sup> \u2013 (1\/2)-3<\/sup>) \u00f7 (1\/4)-3<\/sup><\/strong><\/p>\n\n\n\n

(ii) (32<\/sup> \u2013 22<\/sup>) \u00d7 (2\/3)-3<\/sup><\/strong><\/p>\n\n\n\n

(iii) ((1\/2)-1<\/sup> \u00d7 (-4)-1<\/sup>)-1<\/sup><\/strong><\/p>\n\n\n\n

(iv) (((-1\/4)2<\/sup>)-2<\/sup>)-1<\/sup><\/strong><\/p>\n\n\n\n

(v) ((2\/3)2<\/sup>)3<\/sup> \u00d7 (1\/3)-4<\/sup> \u00d7 3-1<\/sup> \u00d7 6-1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> ((1\/3)-3<\/sup> \u2013 (1\/2)-3<\/sup>) \u00f7 (1\/4)-3<\/sup><\/p>\n\n\n\n

(33<\/sup> \u2013 23<\/sup>) \u00f7 43<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(27-8) \u00f7 64<\/p>\n\n\n\n

19 \u00f7 64<\/p>\n\n\n\n

19 \u00d7 1\/64 (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

19\/64<\/p>\n\n\n\n

(ii)<\/strong> (32<\/sup> \u2013 22<\/sup>) \u00d7 (2\/3)-3<\/sup><\/p>\n\n\n\n

(9 \u2013 4) \u00d7 (3\/2)3<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

5 \u00d7 (27\/8)<\/p>\n\n\n\n

135\/8<\/p>\n\n\n\n

(iii)<\/strong> ((1\/2)-1<\/sup> \u00d7 (-4)-1<\/sup>)-1<\/sup><\/p>\n\n\n\n

(21<\/sup> \u00d7 (1\/-4))-1<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(1\/-2)-1<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

-21<\/sup><\/p>\n\n\n\n

-2<\/p>\n\n\n\n

(iv)<\/strong> (((-1\/4)2<\/sup>)-2<\/sup>)-1<\/sup><\/p>\n\n\n\n

((-1\/16)-2<\/sup>)-1<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

((-16)2<\/sup>)-1<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(256)-1<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

1\/256<\/p>\n\n\n\n

(v)<\/strong> ((2\/3)2<\/sup>)3<\/sup> \u00d7 (1\/3)-4<\/sup> \u00d7 3-1<\/sup> \u00d7 6-1<\/sup><\/p>\n\n\n\n

(4\/9)3<\/sup> \u00d7 34<\/sup> \u00d7 1\/3 \u00d7 1\/6 (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

(64\/729) \u00d7 81 \u00d7 1\/3 \u00d7 1\/6<\/p>\n\n\n\n

(64\/729) \u00d7 27 \u00d7 1\/6<\/p>\n\n\n\n

32\/729 \u00d7 27 \u00d7 1\/3<\/p>\n\n\n\n

32\/729 \u00d7 9<\/p>\n\n\n\n

32\/81<\/p>\n\n\n\n

8. By what number should 5-1<\/sup> be multiplied so that the product may be equal to (-7)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, 5-1<\/sup> \u00d7 x = (-7)-1<\/sup><\/p>\n\n\n\n

1\/5 \u00d7 x = 1\/-7 (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

x = (-1\/7) \/ (1\/5)<\/p>\n\n\n\n

= (-1\/7) \u00d7 (5\/1) (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

= -5\/7<\/p>\n\n\n\n

9. By what number should (1\/2)-1<\/sup> be multiplied so that the product may be equal to (-4\/7)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, (1\/2)-1<\/sup> \u00d7 x = (-4\/7)-1<\/sup><\/p>\n\n\n\n

1\/(1\/2) \u00d7 x = 1\/(-4\/7) (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

x = (-7\/4) \/ (2\/1)<\/p>\n\n\n\n

= (-7\/4) \u00d7 (1\/2) (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

= -7\/8<\/p>\n\n\n\n

10. By what number should (-15)-1<\/sup> be divided so that the quotient may be equal to (-5)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, (-15)-1<\/sup> \u00f7 x = (-5)-1<\/sup> (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

1\/-15 \u00d7 1\/x = 1\/-5 (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

1\/x = (1\u00d7-15)\/-5<\/p>\n\n\n\n

1\/x = 3<\/p>\n\n\n\n

x = 1\/3<\/p>\n\n\n\n

11. By what number should (5\/3)-2<\/sup> be multiplied so that the product may be (7\/3)-1<\/sup>?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider a number x<\/p>\n\n\n\n

So, (5\/3)-2<\/sup> \u00d7 x = (7\/3)-1<\/sup><\/p>\n\n\n\n

1\/(5\/3)2<\/sup> \u00d7 x = 1\/(7\/3) (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

x = (3\/7) \/ (3\/5)2<\/sup><\/p>\n\n\n\n

= (3\/7) \/ (9\/25)<\/p>\n\n\n\n

= (3\/7) \u00d7 (25\/9) (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

= (1\/7) \u00d7 (25\/3)<\/p>\n\n\n\n

= 25\/21<\/p>\n\n\n\n

12. Find x, if<\/strong><\/p>\n\n\n\n

(i) (1\/4)-4<\/sup> \u00d7 (1\/4)-8<\/sup> = (1\/4)-4x<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(1\/4)-4<\/sup> \u00d7 (1\/4)-8<\/sup> = (1\/4)-4x<\/sup><\/p>\n\n\n\n

(1\/4)-4-8<\/sup> = (1\/4)-4x<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

(1\/4)-12<\/sup> = (1\/4)-4x<\/sup><\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

-12 = -4x<\/p>\n\n\n\n

x = -12\/-4<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

(ii) (-1\/2)-19<\/sup> \u00f7 (-1\/2)8<\/sup> = (-1\/2)-2x+1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(-1\/2)-19<\/sup> \u00f7 (-1\/2)8<\/sup> = (-1\/2)-2x+1<\/sup><\/p>\n\n\n\n

(1\/2)-19-8<\/sup> = (1\/2)-2x+1<\/sup> (we know that an<\/sup> \u00f7 am<\/sup> = an-m<\/sup>)<\/p>\n\n\n\n

(1\/2)-27<\/sup> = (1\/2)-2x+1<\/sup><\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

-27 = -2x+1<\/p>\n\n\n\n

-2x = -27-1<\/p>\n\n\n\n

x = -28\/-2<\/p>\n\n\n\n

= 14<\/p>\n\n\n\n

(iii) (3\/2)-3<\/sup> \u00d7 (3\/2)5<\/sup> = (3\/2)2x+1<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(3\/2)-3<\/sup> \u00d7 (3\/2)5<\/sup> = (3\/2)2x+1<\/sup><\/p>\n\n\n\n

(3\/2)-3+5<\/sup> = (3\/2)2x+1<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

(3\/2)2<\/sup> = (3\/2)2x+1<\/sup><\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

2 = 2x+1<\/p>\n\n\n\n

2x = 2-1<\/p>\n\n\n\n

x = 1\/2<\/p>\n\n\n\n

(iv) (2\/5)-3<\/sup> \u00d7 (2\/5)15<\/sup> = (2\/5)2+3x<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(2\/5)-3<\/sup> \u00d7 (2\/5)15<\/sup> = (2\/5)2+3x<\/sup><\/p>\n\n\n\n

(2\/5)-3+15<\/sup> = (2\/5)2+3x<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

(2\/5)12<\/sup> = (2\/5)2+3x<\/sup><\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

12 = 2+3x<\/p>\n\n\n\n

3x = 12-2<\/p>\n\n\n\n

x = 10\/3<\/p>\n\n\n\n

(v) (5\/4)-x<\/sup> \u00f7 (5\/4)-4<\/sup> = (5\/4)5<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(5\/4)-x<\/sup> \u00f7 (5\/4)-4<\/sup> = (5\/4)5<\/sup><\/p>\n\n\n\n

(5\/4)-x+4<\/sup> = (5\/4)5<\/sup> (we know that an<\/sup> \u00f7 am<\/sup> = an-m<\/sup>)<\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

-x+4 = 5<\/p>\n\n\n\n

-x = 5-4<\/p>\n\n\n\n

-x = 1<\/p>\n\n\n\n

x = -1<\/p>\n\n\n\n

(vi) (8\/3)2x+1<\/sup> \u00d7 (8\/3)5<\/sup> = (8\/3) x+2<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(8\/3)2x+1<\/sup> \u00d7 (8\/3)5<\/sup> = (8\/3)x+2<\/sup><\/p>\n\n\n\n

(8\/3)2x+1+5<\/sup> = (8\/3) x+2<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

(8\/3)2x+6<\/sup> = (8\/3) x+2<\/sup><\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

2x+6 = x+2<\/p>\n\n\n\n

2x-x = -6+2<\/p>\n\n\n\n

x = -4<\/p>\n\n\n\n

13. (i) If x= (3\/2)2<\/sup> \u00d7 (2\/3)-4<\/sup>, find the value of x-2<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x= (3\/2)2<\/sup> \u00d7 (2\/3)-4<\/sup><\/p>\n\n\n\n

= (3\/2)2<\/sup> \u00d7 (3\/2)4<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

= (3\/2)2+4<\/sup> (we know that an<\/sup> \u00d7 am<\/sup> = an+m<\/sup>)<\/p>\n\n\n\n

= (3\/2)6<\/sup><\/p>\n\n\n\n

x-2<\/sup> = ((3\/2)6<\/sup>)-2<\/sup><\/p>\n\n\n\n

= (3\/2)-12<\/sup><\/p>\n\n\n\n

= (2\/3)12<\/sup><\/p>\n\n\n\n

(ii) If x = (4\/5)-2<\/sup> \u00f7 (1\/4)2<\/sup>, find the value of x-1<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

x = (4\/5)-2<\/sup> \u00f7<\/strong> (1\/4)2<\/sup><\/p>\n\n\n\n

= (5\/4)2<\/sup> \u00f7 <\/strong>(1\/4)2<\/sup> (we know that 1\/an<\/sup> = a-n<\/sup>)<\/p>\n\n\n\n

= (5\/4)2<\/sup> \u00d7 (4\/1)2<\/sup> (we know that 1\/a \u00f7 1\/b = 1\/a \u00d7 b\/1)<\/p>\n\n\n\n

= 25\/16 \u00d7 16<\/p>\n\n\n\n

= 25<\/p>\n\n\n\n

x-1<\/sup> = 1\/25<\/p>\n\n\n\n

14. Find the value of x for which 52x<\/sup> \u00f7 5-3<\/sup> = 55<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

52x<\/sup> \u00f7 5-3<\/sup> = 55<\/sup><\/p>\n\n\n\n

52x+3<\/sup> = 55<\/sup> (we know that an<\/sup> \u00f7 am<\/sup> = an-m<\/sup>)<\/p>\n\n\n\n

When the bases are same we can directly equate the coefficients<\/p>\n\n\n\n

2x+3 = 5<\/p>\n\n\n\n

2x = 5-3<\/p>\n\n\n\n

2x = 2<\/p>\n\n\n\n

x = 1<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 2.3 PAGE NO: 2.22<\/h4>\n\n\n\n

1. Express the following numbers in standard form:<\/strong><\/p>\n\n\n\n

(i) 6020000000000000<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 6020000000000000 in standard form, count the total digits leaving 1st digit from the left. So the total number of digits becomes the power of 10. Therefore the decimal comes after the 1st digit.<\/p>\n\n\n\n

the total digits leaving 1st digit from the left is 15<\/p>\n\n\n\n

\u2234 the standard form is 6.02 \u00d7 1015<\/sup><\/p>\n\n\n\n

(ii) 0.00000000000942<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 0.00000000000942 in standard form,<\/p>\n\n\n\n

Any number after the decimal point the powers become negative. Total digits after decimal is 12<\/p>\n\n\n\n

\u2234 the standard form is 9.42 \u00d7 10-12<\/sup><\/p>\n\n\n\n

(iii) 0.00000000085<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 0.00000000085 in standard form,<\/p>\n\n\n\n

Any number after the decimal point the powers become negative. Total digits after decimal is 10<\/p>\n\n\n\n

\u2234 the standard form is 8.5 \u00d7 10-10<\/sup><\/p>\n\n\n\n

(iv) 846 \u00d7 107<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 846 \u00d7 107<\/sup> in standard form, count the total digits leaving 1st digit from the left. So the total number of digits becomes the power of 10. Therefore the decimal comes after the 1st digit.<\/p>\n\n\n\n

the total digits leaving 1st digit from the left is 2<\/p>\n\n\n\n

846 \u00d7 107 <\/sup>= 8.46 \u00d7 102 <\/sup>\u00d7 107<\/sup> = 8.46 \u00d7 102+7<\/sup> = 8.46 \u00d7 109<\/sup><\/p>\n\n\n\n

(v) 3759 \u00d7 10-4<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 3759 \u00d7 10-4 <\/sup>in standard form, count the total digits leaving 1st digit from the left. So the total number of digits becomes the power of 10. Therefore the decimal comes after the 1st digit.<\/p>\n\n\n\n

the total digits leaving 1st digit from the left is 3<\/p>\n\n\n\n

3759 \u00d7 10-4 <\/sup>= 3.759 \u00d7 103 <\/sup>\u00d7 10-4<\/sup> = 3.759 \u00d7 103+(-4)<\/sup> = 3.759 \u00d7 10-1<\/sup><\/p>\n\n\n\n

(vi) 0.00072984<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

To express 0.00072984 in standard form,<\/p>\n\n\n\n

Any number after the decimal point the powers become negative. Total digits after decimal is 4<\/p>\n\n\n\n

\u2234 the standard form is 7.2984 \u00d7 10-4<\/sup><\/p>\n\n\n\n

(vii) 0.000437 \u00d7 104<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express 0.000437 \u00d7 104 <\/sup>in standard form,<\/p>\n\n\n\n

Any number after the decimal point the powers become negative. Total digits after decimal is 4<\/p>\n\n\n\n

\u2234 the standard form is 4.37 \u00d7 10-4<\/sup> \u00d7 104<\/sup> = 4.37<\/p>\n\n\n\n

(viii) 4 \u00f7 100000<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

To express in standard form count the number of zeros of the divisor. This count becomes the negative power of 10.<\/p>\n\n\n\n

\u2234 the standard form is 4 \u00d7 10-5<\/sup><\/p>\n\n\n\n

2. Write the following numbers in the usual form:<\/strong><\/p>\n\n\n\n

(i) 4.83 \u00d7 107<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are positive the usual form of number is written after the multiplication of the given number, then place the decimal point after counting from right.<\/p>\n\n\n\n

4.83 \u00d7 10000000 = 4830000000<\/p>\n\n\n\n

48300000.00<\/p>\n\n\n\n

\u2234 the usual form is 48300000<\/p>\n\n\n\n

(ii) 3.02 \u00d7 10-6<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are negative the decimal is placed to the left of the number.<\/p>\n\n\n\n

3.02 \u00d7 10-6<\/sup> here, the power is -6, so the decimal shifts 6 places to left.<\/p>\n\n\n\n

\u2234 the usual form is 0.00000302<\/p>\n\n\n\n

(iii) 4.5 \u00d7 104<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are positive the usual form of number is written after the multiplication of the given number, then place the decimal point after counting from right.<\/p>\n\n\n\n

4.5 \u00d7 10000 = 450000<\/p>\n\n\n\n

45000.0<\/p>\n\n\n\n

\u2234 the usual form is 45000<\/p>\n\n\n\n

(iv) 3 \u00d7 10-8<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are negative the decimal is placed to the left of the number.<\/p>\n\n\n\n

3 \u00d7 10-6<\/sup> here, the power is -8, so the decimal shifts 8 places to left.<\/p>\n\n\n\n

\u2234 the usual form is 0.00000003<\/p>\n\n\n\n

(v) 1.0001 \u00d7 109<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are positive the usual form of number is written after the multiplication of the given number, then place the decimal point after counting from right.<\/p>\n\n\n\n

1.0001 \u00d7 1000000000 = 10001000000000<\/p>\n\n\n\n

1000100000.0000<\/p>\n\n\n\n

\u2234 the usual form is 1000100000<\/p>\n\n\n\n

(vi) 5.8 \u00d7 102<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are positive the usual form of number is written after the multiplication of the given number, then place the decimal point after counting from right.<\/p>\n\n\n\n

5.8 \u00d7 100 = 5800<\/p>\n\n\n\n

580.0<\/p>\n\n\n\n

\u2234 the usual form is 580<\/p>\n\n\n\n

(vii) 3.61492 \u00d7 106<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are positive the usual form of number is written after the multiplication of the given number, then place the decimal point after counting from right.<\/p>\n\n\n\n

3.61492 \u00d7 1000000 = 361492000000<\/p>\n\n\n\n

3614920.00000<\/p>\n\n\n\n

\u2234 the usual form is 3614920<\/p>\n\n\n\n

(vii) 3.25 \u00d7 10-7<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

When the powers are negative the decimal is placed to the left of the number.<\/p>\n\n\n\n

3.25 \u00d7 10-7<\/sup> here, the power is -7, so the decimal shifts 7 places to left.<\/p>\n\n\n\n

\u2234 the usual form is 0.000000325<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 2\u2013Powers<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 2\u2013Powers PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n