Simplify using laws of exponents :
1) 54 x 57 2) d2 x d6 3) k8 ÷ k3 4) 75 ÷ 76 5) (x2)5 6) (34)4 7) p3 ÷ p7 8) n9 x n3 9) (5t)3 10) 7x · 72 |
11) 103 ÷ 10q 12) (c4)m 13) (6x2) (4x2) 14) (3x3y2)(-6 y5) 15) (5p3)(-m8 p2) 16) (10g3 h8 v6) (11g h8) 17) (4 f9 h3) (-5 f6) (-3 h2) 18) (-22 x3 y4) ((-3)2 x4 y4) 19) (3 xa yb zc )(-yf zg ) 20) (x2 y)4 |
Problem 1 :
54 x 57
Solution :
= 54 x 57
Since the bases are same, we have to use only one base and add the powers.
= 54+7
= 511
Problem 2 :
d2 x d6
Solution :
= d2 x d6
Since the bases are same, we have to use only one base and add the powers.
= d2 + 6
= d8
Problem 3 :
k8 ÷ k3
Solution :
= k8 ÷ k3
Since the bases are same, we have to use only one base and subtract the powers.
= k8 - 3
= k5
Problem 4 :
75 ÷ 76
Solution :
= 75 ÷ 76
Since the bases are same, we have to use only one base and subtract the powers.
= 75 - 6
= 7-1
By changing the negative exponent as positive, we have to change flip the base.
= (1/7)1
= 1/7
So, the answer is 1/7.
Problem 5 :
(x2)5
Solution :
= (x2)5
Since we have power raised by another power, we have to multiply the powers.
= x2·5
= x10
So, the answer is x10.
Problem 6 :
(34)4
Solution :
= (34)4
Since we have power raised by another power, we have to multiply the powers.
= 34·4
= 316
So, the answer is 316.
Problem 7 :
p3 ÷ p7
Solution :
= p3 ÷ p7
Since the bases are same, we have to use only one base and subtract the powers.
= p3 - 7
= p-4
By changing the negative exponent as positive, we have to change flip the base.
= (1/p)4
= 1/p4
So, the answer is 1/p4.
Problem 8 :
n9 x n3
Solution :
= n9 x n3
Since the bases are same, we have to use only one base and add the powers.
= n3 + 9
= n12
Problem 9 :
(5t)3
Solution :
= (5t)3
Since we have power raised by another power, we have to multiply the powers.
= 5t(3)
= 53t
Problem 10 :
7x · 72
Solution :
= 7x · 72
Since bases are multiplied, we have to put the base once and add the powers.
= 7(x + 2)
So, the answer is 7(x + 2).
Problem 11 :
103 ÷ 10q
Solution :
= 103 ÷ 10q
Since bases are divided, we have to put only one base and subtract the powers.
= 103 - q
Problem 12 :
(c4)m
Solution :
= (c4)m
Since we have power raised by another power, we have to multiply the powers.
= c4m
Problem 13 :
(6x2) (4x2)
Solution :
= (6x2) (4x2)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 6 (4) x2x2
= 24 x2 + 2
= 24 x4
Problem 14 :
(3x3y2)(-6 y5)
Solution :
= (3x3y2)(-6 y5)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 3 (6) x3 y2 y5
= 24 x2 + 2
= 24 x4
= 24 x4
Problem 15 :
(5p3)(-m8 p2)
Solution :
= (5p3)(-m8 p2)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 5(-1) p3p2m8
= -5 p2+3m8
= -5 p5m8
Problem 16 :
(10g3 h8 v6) (11g h8)
Solution :
= (10g3 h8 v6) (11g h8)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 10 (11) g3 · g h8· h8 v6
= 110 g(3 + 1) h(8+8)v6
= 110 g4 h16v6
Problem 17 :
(4 f9 h3) (-5 f6) (-3 h2)
Solution :
= (4 f9 h3) (-5 f6) (-3 h2)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 4(-5)(-3) f9 · f6 h3 · h2
= -60 f9+6 h3+2
= -60 f15 h5
Problem 18 :
(-22 x3 y4)((-3)2 x4 y4)
Solution :
= (-22 x3 y4)((-3)2 x4 y4)
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 4(-3)2x3 · x4 y3 · y4
= 36 x3+4 y3+4
= 36 x7 y7
Problem 19 :
(3 xa yb zc )(-yf zg )
Solution :
= (3 xa yb zc )(-yf zg )
First we have to multiply signs, then multiply the coefficients. Finally multiply the variables.
= 3(-1) xa yb zczg
= -3 xa yb zc+g
Problem 20 :
(x2 y)4
Solution :
= (x2 y)4
Here we have to distribute the power for all the terms which are inside the bracket.
= (x2)4 y4
Since we have power raised by another power, we have to multiply the powers.
= x2(4) y4
= x8 y4
20)
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM