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Instead of writing the repeating the same, we can use the exponents
Simplify each of the following.
Problem 1 :
a โ a2 โ a3
Solution :
= a โ a2 โ a3
Since the base is same for all three terms, we use only one base and add the powers.
= a(1+2+3)
= a6
Problem 2 :
(2 a2 b) โ (4ab2)
Solution :
= (2 a2 b) โ (4ab2)
Multiplying the coefficients = 2 x 4 ==> 8
Multiplying a related terms = a2โ a = a3
Multiplying b related terms = b โ b2 = b3
Since the base is same for all three terms, we use only one base and add the powers.
= 8a3b3
Problem 3 :
(6x2) โ (-3x5)
Solution :
= (6x2) โ (-3x5)
= -18 x2+5
= -18x7
Problem 3 :
(6x2) โ (-3x5)
Solution :
= (6x2) โ (-3x5)
= -18 x2+5
= -18x7
Problem 4 :
(2x2 y3)2
Solution :
= (2x2 y3)2
Distributing the power, we get
= 22 (x2)2 (y3)2
= 22 x4 y6
= 4x4 y6
Problem 5 :
Solution :
(-4)2 and -42 are same ?
No, there is a difference between (-4)2 and -42.
In (-4)2, order of operations (PEMDAS) says to take first.
(-4)2 = (-4) โ (-4)
(-4)2 = 16
Without parentheses, exponents take precedence :
-42 = -4 โ 4
-42 = -16
Sometimes, the result will be same as in (-2)3 and -23.
For a negative number with odd exponent, the result is always negative.
Problem 6 :
Find the value of (i) 4-3 (ii) 1/2-3 (iii) (-2)5 x (-2)-3 (iv) 32/3-2.
Solution :
(i) 4-3 :
= 1/43
= 1/(4 x 4 x 4)
= 1/64
(ii) 1/2-3 :
= 23
= 2 x 2 x 2
= 8
(iii) (-2)5 x (-2)-3 :
= (-2)5 - 3
= (-2)2
= -2 x -2
= 4
(iv) 32/3-2 :
= 32/3-2
= 32 x 32
= 9 x 9
Problem 7 :
Simplify and write the answer in exponential form:
(i) (35 รท 38)5 x 3-5 (ii) (-3)4 x (5/3)4
Solution :
(i) (35 รท 38)5 x 3-5 :
= (35 - 8)5 x 3-5
= (3-3)5 x 3-5
= 3-3 x 5 x 3-5
= 3-15 x 3-5
= 3-15 - 5
= = 3-20
(ii) (-3)4 x (5/3)4 :
= 34 x 54/34
= 54
Problem 8 :
Find x so that (-7)x + 2 x (-7)5 = (-7)10.
Solution :
(-7)x + 2 x (-7)5 = (-7)10
(-7)x + 2 + 5 = (-7)10
(-7)x + 7 = (-7)10
Since the bases are equal, we can equate the exponents.
x + 7 = 10
Subtract 7 from each side.
x = 3
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May 21, 24 08:51 PM
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