INTRODUCTION
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Application of the formulae (i) am ´ an = am + n (ii) am ¸ an = am- n (iii) abmn = bamn (iv) –abmn = –ba+mn (v) am ¸ b-n = am ´ bn |
Example 4: Find the value of –121623
Solution: Applying the above formula (v), we have
–1216–23 = -21623 = -6323 = -62 = 36
Example 5: Find the value of -2-2-2
Solution: Applying the above formula (v), we have
-2-2-2 = –12+2-2 = 14-2 = 42 = 16
Example 6: Find the value of 3432 ¸ 34343
Solution: Applying the above formula (ii), we have
3432 ¸ 34343 = 3432-43 = 34323
= 7323 = 72 = 49
Trick1:
Example: Solve 353
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For any positive integer ‘n’ and a positive rational number ‘a’, nan = a. |
Example: 353 = 5 |
Example 1: Simplify the following:364
Solution: 364 =343
Example 2 : Find the value of x in each of the following:
(i) 34x-7-5 = 0
(ii) 43x+1 = 2
Solution: (i) 34x-7-5 = 0
⟹ 34x-7-5 ⟹ 34x-73 = 53
⟹ 4x – 7 = 125 ∵ nan=a
⟹ x = 33
(ii) 43x+1 = 2
⟹ 43x+14 = 24 ⇒ 3x + 1 = 16
⇒ 3x = 15 ∵ nan=a
⇒ x = 5
Trick2
Example: Solve 33 . 34
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If ‘n’ is a positive integer and ‘a’, ‘b’ are rational numbers, then na nb = nab |
Example: 33 . 34 = 33 ´ 4 = 312 |
Example: Simplify the following 3128
Solution: 3128 = 364 ´ 2 = 364 32
= 343 . 32 = 4 32 [Using 1st Law 343 = 4]
Trick3
Example: Solve 3827
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If ‘n’ is a positive integer and ‘a’, ‘b’ are rational numbers, then nanb =nab . |
Example: 3827 = 38327 = 322333 = 23 |
Example: Simplify each of the following:
(i) 43888448
Solution: (i) 43888448 = 4388848
= 481 = 434 =3 ∵ nan=a
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If ‘m’, ‘n’ are positive integers and ‘a’ is a positive rational number, then mna = mna = nma . |
Ex: 433 = 123 |
Example: Simplify each of the following
(i) 235
Solution: (i) 235 = 65
Trick 4
Example: Evaluate Ex: 54234
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If ‘m’, ‘n’ are positive integers and ‘a’ is positive rational number, then nmapm = nap = mnapm |
Ex: 54234 = 523 = 58 .
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Trick5:
Example: Which surd is larger 33 or 45 ?
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Comparison of Surds of Distinct Orders |
Example: Which surd is larger 33 or 45 ? Solution: The orders of the given surds are 3 and 4 respectively. Now, LCM of 3 and 4 = 12. So, we convert each surd into a surd of order 12. Now, 33 = 1234 = 1281 And 45 = 1253 = 12125 Clearly, 125 > 81 ∴ 12125 > 1281 ⇒ 45 > 33 .
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Example 2: Which is greater 3 or410 .
Solution: The orders of the given surds are 2 and 4 respectively. LCM of 2 and 4 is 4.
So, we convert each surd into a surd of order 4.
Now, 3 = 432 = 49 .
∵ nan=a
Clearly, 10 > 9
∴ 410 > 49 ⇒410 > 3
Example 3: Which is greater 36 or 48 ?
Solution: The orders of the given surds are 3 and 4 respectively. LCM of 3 and 4 is 12.
Convert each surd into a surd of order 12.
Now, 36 = 1264 = 121296
and, 48 = 1283 = 12512
Clearly, 1296 > 512
∴ 121296 > 12572 ⇒ 36 > 48
Example 4: Which is greater 1212 or 2313 ?
Solution: The orders of the given surds are 2 and 3 respectively. LCM of 2 and 3 is 6.
Covert each surd into a surd of order 6 as given below
1212 = 12 = 6123 = 618
And 2313 = 323 = 6232 = 649
Now, 49 > 18 ∵4 ´ 8>9 ´ 1
∴ 649 > 618 ⇒ 2313 > 1212 .
Trick6
Example: Write surds in ascending order 43 , 67 , 1248
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Rule: – Make power same of Step 1:- find LCM of powers p, q, and r = ( L say) Step2:- Divide LCM L from power and make it power of a , b and c
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Example: 43 , 67 , 1248
Step 1:- LCM of 4, 6, and 12 = 12
Step 2:-
43 = 1233 = 1227
67 = 1272 = 1249
448 = 12481 = 1248
Here, 1227 < 1248 < 1249
Therefore
43 < 1248 < 67 |
Trick7
Example: Solve 13+ 2
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Solve 1a+ b
Rule:- if a = b + 1
1a + b = a – b
And
1a – b = a + b
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Example: Solve 13+ 2
Detail Method
Solve by rationalizing
13+ 2 x 3– 23 – 2 = 3– 23-2 = 3 – 2
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Trick :- 13+ 2 = 3 – 2 |
Do Yourself
1. Example: Solve 1112+
111
2. Example Solve 11234
– 1233
Trick 8
Example: Solve 3 + 23 – 2
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Solve: a + ba – b Rule:-
Case 1:- If a = b + 1
a + ba – b = (a+ b)2
Or
a + ba – b = a + b + 2ab
And
Case 2:- if a ≠ b + 1
a + ba – b = a+ba-b + 2aba-b
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Solve: 3 + 23 – 2
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Method 1:- Case 1:- If a = b + 1
3 + 23 – 2 = (3+ 2)2
Or
3 + 23 – 2 = 3 + 2 + 22 x 3
= 5 + 26
Case 2:- if a ≠ b + 1
3 + 23 – 2 = 3+23-2 + 23 x 23-2
= 5 + 26 |
Example1: Solve 34+ 3344 – 33
Solution: On rationalizing
34+ 3344– 33 x 44+ 3344+ 33 = 34 44+ 33 + 33(44+ 33 ) 44-33
= 1496+ 1122 + (1452+ 1089 ) 44-33 = 1496+ 1122 + (1452+ 1089 ) 1
Do Yourself
1. Ex:- Solve 23+2223 – 22
2. Ex:- Solve 34+ 3344 – 33
Trick 9
Example: Solve 3 – 23+ 2
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Rule:-
Case 1:- If a = b + 1
a – ba+ b = (a–b)2
Or a – ba+ b = a + b – 2ab
And
Case 2:- if a ≠ b + 1
a – ba + b = a+ba-b – 2aba-b
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Detail Method:
3 – 23+ 2
On rationalizing
3– 23 + 2 x 3– 23– 2 = (3–2 )23-2 = (3–2 )2
Or
= 3 + 2 – 22 x 3
Or
= 5 – 26
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Method 1:-
Case 1:- If a = b + 1
3 – 23+ 2 = (3–2)2
Or
3 – 23 + 2 = 3 + 2 – 22 x 3
= 5 + 26
Case 2:- if a ≠ b + 1
3 – 23 + 2 = 3+23-2 – 23 x 23-2
= 5 – 26
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Example1: Solve 3 + 23 – 2
Solution:
3 + 23 – 2 =
(3–2)2 = 3–2
Do Yourself
1. Solve 23 –2223 + 22
2. Solve 34 – 3344 + 33
Trick 10
Example: Find value of 7+43
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Rule:- a+2mn = m+n+2mn
= m2+ n2+2mn
= (m+ n)2 = ( m+n ) |
Detail Method :- 7+43
= 4+3+43
= 42+ 32+2 X 23
=(42+ 32)2
= (2 + 3 )
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Example1: Find value of 13-43
Solution: (12– 1)2
( 12–1 ) = 23 – 1
Example2: Solve 28-103
Solution: 28-103 = (25– 3)2 = (5 – 3 )
Example3: Solve:- –3+ 3+87+43
Solution: –3+ 3+87+43
= –3+ 3+8(2+ 3 ) = –3+ 3+16+ 83 )
= –3+ 19 + 83 ) = –3+ 19 + 83 )
= –3+ 4+ 3 = 2
Do Yourself
1. Find value of (28-10 3
2. Find value of 7+4 3
Trick11
Example: Solve 13+4 + 14+5 + 15+6 …………….199+100
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Rule1:– 1a+a+1 + 1a+1+a+2 + 1a+2+a+3 ………….1a+n +a+ n +1
Step 1:- Change the order (bigger term will come 1st)
1a+1+a + 1a+2+a+1 + 1a+3+a+2 ………….1a+n+1 +a+ n
Required answer = ( – 1st term + last term)\
= – a + a+n+1
Or
Rule1:– 1a+a+1 + 1a+1+a+2 + 1a+2+a+3 ………….1a+n +a+ n
without changing the Required answer = (1st term + last term)
= a + a+n+1 |
Trick: 13+4 + 14+5 + 15+6 ….199+100
Step 1:- Change order of terms
14+3 + 15+4 + 16+5 …….1100+99
Required answer = – 3 + 100 = 10 –3 |
Detail Method:-
After rationalization of each term
13+4 x + 14+5 + 15+6 ………….199+100
= 13+4 x3 – 43 – 4 + 14+5x 4– 54 – 5 + 15+6 x5– 65 – 6 ………….199+100x99– 10099 – 100
=3 – 4 + 4– 5 + 5– 6 +………99– 100
= from 2nd term each alternate term will be cancel
= 3– 100 = 3 – 10
Trick12
Example: Solve 110+110+ 110 +… …………..∞
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Required answer =( n +1 ) Note:- a ≠ n (n + 1), not possible to break in n ( n + 1 ), then Required answer = cannot determine or imaginary solution 0r not defined |
Shortcut Method 1 110+110+ 110 +… ….∞ Step 1:- 110 = 10 x 11 Since + ve sign in question Required answer = 11 |
Shortcut Method 2 Step1 :- check nearest complete square 100 or 121 Since 121 > 100 Or 112 > 102 Therefore required answer = 11 |
Detail Method:
110+110+ 110 +… ….∞
Let X = 110+110+ 110 +… ….∞
ð x = 110+x
ð squaring both side
ð x2 = 110 + x = 0
ð x2 – x -110 = 0
ð (x – 11) (x +10) = 0
ð x = 11 , x = – 10
Example1: Solve 71+71+ 71 +… …………..∞
Solution: Here 71 is not breakable in consecutive factor, n (n + 1)
Required answer = cannot determine
Do Yourself
a) Solve 42+42+ 42 +… …………..∞
b) Solve 71+71+ 71 +… …………..∞
Trick13
Example: Solve 110-110- 110-… …………..∞
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Rule:- a-a- a-… Step 1:- Write a = n(n + 1)
Since – ve sign in question
Required answer = n
Note:- a ≠ n (n + 1), not possible to break in n ( n + 1 ), then
Required answer = cannot determine or imaginary solution 0r not defined |
Shortcut 1: 110-110- 110-… ∞
Step 1:- 110 = 10 x 11
Since – ve sign in question
Required answer = 10 |
Shortcut 2 Step1 :- check nearest complete square
100 or 121
Since 121 < 100
Or 112 < 102
Therefore required answer = 10 |
Detail Method:
Let X = 110-110- 110-… ….∞
ð x = 110-x
ð squaring both side
ð x2 = 110 – x = 0
ð x2 + x -110 = 0
ð (x + 11) (x -10) = 0
ð x = -11 , x = + 10
ð Required ans.= 10
Trick14
Example: Solve 110 110 110 … …………..∞
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Rule:- a a a … …………..∞
Required answer = x |
Shortcut Method :- 110 110 Required answer = x = 110 |
Detail method:-
Let x = 110 110
110 … …………..∞
ð x = 110 x
Squaring both side
x2 = 110 x
ð x = 110
Example1: 2 x 34 x 2 x 34 x ……………..∞
Solution: Let y = 2 x 34 x 2 x 34 x ……………..∞
ð y = 2 x 34 x y
Squaring both side
y2 =2 x 34 y
Cubic both side
(y2)3 = 8 x 4y or y6 = 8 x 4y
ð y5 = 32 => y = 2
Trick15
Example: Solve 110 110 110 … …………..110
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Rule:- a1 a1 Required answer = a2n– 12n |
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Method1 :- 110 x110 x 110 x Required answer = 11023– 123 = 11078 |
Do Yourself
a) 11 11 11 … …………..11
b) 10 10 10 … …………..10
c) 1000 1000 1000 … …………..1000
Trick16
Example: Find the range 1+ 292
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Rule How to find Range 1+ m2
Step 1:- consider any digit which either less or greater than m and having perfect square root
Step2:- Required range
1+ a2 < 1+ m2 <1+ b2 |
Shortcut Method 1+ 292
Here 25 < 29 < 36
Required range
1+ 252 < 1+ m2 < 1+ 362
∴ 3 < m <3.5 |
Do Yourself
a) Find Range of
b) Find Range of
Trick17
Example: If x = 7 + 4 3 find 1x
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Rule If the difference of square of two number is 1, sign of their inverse(conjugate) number will be change
Step1:- If x = a + b c
Step2 :- 1x = a – b c |
Shortcut Method
Step1:- If x = 7 + 4 3
Here ( 7 ) 2 – (4 3 )2
= 49 – 48 = 1 ∴ Step2 :- 1x = 7 – 4 3
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Example1: If x = 17+43 and y = 17-
43 then find 1x+1 + 1y+1
Solution: If x = 17+43 Then, 1x = 17- 43
∴ y = 1x
Now, 1x+1 + 1y+1= 1x+1 + 11x+1 = 1x+1 + xx+1 = x + 1x+1 = 1
Example2: If x = (3+ 2)-3
, y = (3– 2)-3
, find 1x+1 + 1y+1
Solution: If x = (3+ 2)-3 then 1x = (3– 2)-3
∴ y = 1x
Now, 1x+1 + 1y+1= 1x+1 + 11x+1 = 1x+1 + xx+1 = x + 1x+1 = 1
Example3: If x = 7 + 4 3 find x + 1x
Solution: x = 7 + 4 3 then,
1x = 7 – 4 3
∴ x + 1x = 7 + 4 3 + 7 – 4 3 = 14
Trick18
Example: If x = 14 find the value of x5 -15x4 + 15x3 – 15x2 + 15x
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Rule If x = a then the value of 1x5 -bx4 + bx3 – bx2 + bx Step1:- Break coefficients b = a + 1 Step2:-All term will be cancel out , and then put value of x = a |
Shortcut Method If x = 14 find the value of x5 -15x4 + 15x3 – 15x2 + 15x Step 1:- 15 = 14 + 1 (x5 -14x4 – x4 + 14x3 + x3 – 14x2 – x2 + 14x) + x
ð x = 14 ( Put x = 14 and other term will be cancel out |
Example1: If x = 12 find the value of x6– 13x5 + 13x4 – 13x3 + 15x2 – 13x +5
Solution: x6– 13x5 + 13x4 – 13x3 + 15x2 – 13x +5
= x6– 13x5 + 13x4 – 13x3 + 13x2 + 2 x2 – 13x +5
= (x6– 13x5 + 13x4 – 13x3 + 13x2 – 13x) + (5 + 2 x2)
Here, (x6– 13x5 + 13x4 – 13x3 + 13x2 – 13x)= 0 (put x =12)
∴ 5 + 2 x2 = 5 + 2 (12)2 = 193
Trick19
Example: Find value of 33x-8
–
4=0
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Shortcut Method
1:- 33x-8 – 4 =0
Solve by option
Let we have option 12, 24, 8 and 36
Try one by one with oral calculation
Let x = 24
33x 24-8 – 4
= 364 – 4 = 0
Hence satisfied the result
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Method2:- 33x-8 – 4=0
ð 33x-8 = 4
Cubic both side
3x – 8 = (4)3
ð 3x – 8 = 64
ð 3x = 72
x = 24 |
Do Yourself
a) Find value of 44x+ 8 – 3=0
b) Find value of 34x-7 – 5=0
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Some More Example |
Example1: What should come in place of both x in the equation x128 =
162x
Solution: Let x128 =
162x
Then, x2 =128 x 162
=64 x 2 x
18 x 9
=826232
= 8 x 6 x 3 = 144.
x = 144 = 12.
Example2: The least perfect square, which is divisible by each of 21, 36 and 66 is:
Solution: L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444
Example3: If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?
Solution: 35 + 125 = 17.88
35 + 25x5 = 17.88
35 + 55 = 17.88
85 = 17.88
5 = 2.235
80 + 65 = 16 x 5 + 65
= 45 + 65
= 105 = (10 x 2.235) = 22.35
Example4: If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is:
Solution: 4a2 – 4a + 1 + 3a = (1)2 + (2a)2 – 2 x 1 x 2a + 3a
=(1 – 2a)2 + 3a
= (1 – 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
Example5: If x = 3+13-1 and y = 3-13+1 , then the value of (x2 + y2) is:
Solution: x =3+13-1 x 3+13+1
= (3+1)23-1 = 3+1+232 = 2 +3 .
y =3–13+1 x3–13–1 = (3–1)23-1 =3+1-232 = 2 – 3
x2 + y2 = (2 +3 )2 + (2 – 3 )2
= 2(4 + 3) = 14
Example6: A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
Solution: Money collected = (59.29 x 100) paise = 5929 paise.
Number of members = 5929 = 77.
Example7: The square root of (7 + 35 ) (7 – 35 ) is
Solution: (7 + 35) (7 – 35) =(7)2 –
(35)2 = 49 – 45 =4= 2
Example8: If 5 = 2.236, then the value of
52–105 + 125 is equal to:
Solution: 52–105 + 125 =
(5)2-20+25 x
5525
= 5-20+5025 = 3525x55 = 35510
= 7 x 2.2362
= 7 x 1.118 = 7.826
Example9: 62511x1425x11196 is equal to:
Solution: Given Expression = 2511 x
145 x
1114 = 5.
Example10: 0.0169 x? = 1.3
Solution: Let 0.0169 x? = 1.3
Then, 0.0169x = (1.3)2 = 1.69
x = 1.69 = 100
Example11: 3–132 Simplifies to:
Solution: 3–132 = (3 )2 + 132 – 2 x 3 x13
= 3 + 13 – 2 = 1 + 13
= 43
Example12: (17)3.5 x (17)? = 178
Solution: Let (17)3.5 x (17)x = 178.
Then, (17)3.5 + x = 178.
3.5 + x = 8
x = (8 – 3.5)
x = 4.5
Example13: Ifabx-1=bax-3 , then the value of x is:
Solution: Given abx-1=bax-3
abx-1=abx-3 =ab3-x
x – 1 = 3 – x
2x = 4
x = 2.
Example14: Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
Solution: xz = y2 10(0.48z) = 10(2 x 0.70) = 101.40
0.48z = 1.40
z = 14048 = 3512 = 2.9 (approx.)
Example15: If 5a = 3125, then the value of 5(a – 3) is:
Solution: 5a = 3125 5a = 55
a = 5.
5(a – 3) = 5(5 – 3) = 52 = 25.
Example16: If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:
Solution: 3x – y = 27 = 33 x – y = 3 ……… (i)
3x + y = 243 = 35 x + y = 5 …….. (ii)
On solving (i) and (ii), we get x = 4.
Example17: (256)0.16 x (256)0.09 = ?
(256)0.16 x (256)0.09 = (256)(0.16 + 0.09)
= (256)0.25 = (256)(25/100) = (256)(1/4) = (44)(1/4)
= 44(1/4) = 41 = 4
Example18: The value of [(10)150 ÷ (10)146]
Solution: (10)150 ÷ (10)146 = (10)150(10)146
= 10150 – 146 = 104 = 10000
Example19: 11+x(b–a)+x(c–a) +11+x(a–b)+x(c–b)+11+x(b–c)+x(a–c) = ?
Solution: Given Exp. = 11+xbxa+xcxa + 11+xaxb+xcxb + 11+xbxc+xaxc
= xa(xa+xb+xc) + xb(xa+xb+xc) + xc(xa+xb+xc)
= (xa+xb+xc)(xa+xb+xc) = 1.
Example20: (25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
:
Solution: Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x.
Then, (52)7.5 x (5)2.5(53)1.5 = 5x
5(2x7.5)5(3x1.5) = 5x
51.5x52.554.5 = 5x
5x = 5(15 + 2.5 – 4.5)
ð 5x = 513
x = 13.
Example21: (0.04)-1.5 =?
Solution: (0.04)-1.5 = (25)(3/2)
= (52)(3/2) = (5)2 x (3/2) = 53 = 125.
Example22: 243n5 x 32n+19n x3n-1 = ?
Solution: Given Expression
243n5 x 32n+19n x3n–1
= 35n5x32n+1(32)nx3n-1
=3nx32n+1(3)2nx 3n-1 = 3(n+2n+1)3(2n+n-1) = 33n+133n-1
= 33n+1-3n+1 = 32 = 9.
Example23: 11 + a(n-m) + 11+a(m-n) = ?
Solution: 11 + a(n–m) + 11+a(m–n) = 11+anam
+ 11+aman
= am(am+an) +an(am+an) = (am+an)am+an = 1.
Example24: If m and n are whole numbers such that m n = 121, the value of (m – 1) n + 1 is:
Solution: We know that 112 = 121.
Putting m = 11 and n = 2, we get:
(m – 1)n + 1 = (11 – 1)(2 + 1) = 103 = 1000.
Example25: xbxcb+c-a . xcxac+a-b. xaxba+b-c = ?
Solution: Given Exp.
= xb-cb+c-a . xc-ac+a-b . xa-ba+b-c
= xb-cb+c-ab-c . xc-ac+a-bc-a . xa-ba+b-ca-b
= xb2–c2+ c2–a2+ a2– b2 . x–b-c-bc-a-ca-b
= x0 X x0 = (1 x 1) = 1.
Example26: If x = 3 + 22 , then the value of x–1x is:
Solution: x–1x2 = x + 1x – 2
= (3 + 22 ) +
1(3+22) – 2
= (3 + 22 ) + 1(3+22) X (3-22)(3-22) – 2
= (3 + 22 ) + (3 – 22 ) – 2
= x–1x2 = 4
x–1x = 2
