Instead of writing the repeating the same, we can use the exponents
Simplify each of the following.
Problem 1 :
a ⋅ a^{2} ⋅ a^{3}
Solution :
= a ⋅ a^{2} ⋅ a^{3}
Since the base is same for all three terms, we use only one base and add the powers.
= a^{(1+2+3)}
= a^{6}
Problem 2 :
(2 a^{2} b) ⋅ (4ab^{2})
Solution :
= (2 a^{2} b) ⋅ (4ab^{2})
Multiplying the coefficients = 2 x 4 ==> 8
Multiplying a related terms = a^{2}⋅ a = a^{3}
Multiplying b related terms = b ⋅ b^{2} = b^{3}
Since the base is same for all three terms, we use only one base and add the powers.
= 8a^{3}b^{3}
Problem 3 :
(6x^{2}) ⋅ (-3x^{5})
Solution :
= (6x^{2}) ⋅ (-3x^{5})
= -18 x^{2+5}
= -18x^{7}
Problem 3 :
(6x^{2}) ⋅ (-3x^{5})
Solution :
= (6x^{2}) ⋅ (-3x^{5})
= -18 x^{2+5}
= -18x^{7}
Problem 4 :
(2x^{2 }y^{3})^{2}
Solution :
= (2x^{2 }y^{3})^{2}
Distributing the power, we get
= 2^{2} (x^{2})^{2 } (y^{3})^{2}
= 2^{2} x^{4}^{ }y^{6}
= 4x^{4}^{ }y^{6}
Problem 5 :
Solution :
(-4)^{2} and -4^{2 }are same ?
No, there is a difference between (-4)^{2} and -4^{2}.
In (-4)^{2}, order of operations (PEMDAS) says to take first.
(-4)^{2 }= (-4) ⋅ (-4)
(-4)^{2 }= 16
Without parentheses, exponents take precedence :
-4^{2 }= -4 ⋅ 4
-4^{2 }= -16
Sometimes, the result will be same as in (-2)^{3} and -2^{3}.
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) 3^{2}/3^{-2}.
Solution :
(i) 4^{-3 }:
= 1/4^{3}
= 1/(4 x 4 x 4)
= 1/64
(ii) 1/2^{-3 }:
= 2^{3}
= 2 x 2 x 2
= 8
(iii) (-2)^{5} x (-2)^{-3}^{ }:
= (-2)^{5 - 3}
= (-2)^{2}
= -2 x -2
= 4
(iv) 3^{2}/3^{-2}^{ }:
= 3^{2}/3^{-2}
= 3^{2} x 3^{2}
= 9 x 9
Problem 7 :
Simplify and write the answer in exponential form:
(i) (3^{5 }÷ 3^{8})^{5} x 3^{-5} (ii) (-3)^{4 }x (5/3)^{4}
Solution :
(i) (3^{5 }÷ 3^{8})^{5} x 3^{-5 }:
= (3^{5 - }^{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}^{ }:
= 3^{4 }x 5^{4}/3^{4}
= 5^{4}
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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