To evaluate an expression with exponents, we should to be aware of the rules of exponents.
The Required Rules of Exponents are below.
Example 1 :
yx^{1/3 }⋅ xy^{3/2}
Solution :
= yx^{1/3 }⋅ xy^{3/2}
By writing each term separately, we get
= y ⋅ x^{1/3 }⋅ x ⋅ y^{3/2}
By combining the like terms, we get
= x^{1/3 }⋅ x^{ }⋅ y^{3/2 }⋅ y
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= x^{1/3+1 }⋅ y^{3/2+1}
= x^{4/3 }⋅ y^{5}^{/2}
So, the answer is x^{4/3 }⋅ y^{5}^{/2}
Example 2 :
4v^{2/3 }⋅ v^{-1}
Solution :
= 4v^{2/3 }⋅ v^{-1}
By writing each term separately, we get
= 4 ⋅ v^{2/3 }⋅ v^{-1}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= 4 ⋅ v^{2/3}^{ - 1}
= 4v^{-1/3}
By using the negative exponent a^{-x }= 1/a^{x}
= (4)/v^{1/3}
So, the answer is (4)/v^{1/3}.
Example 3 :
(a^{1/2 }b^{1/2})^{-1}
Solution :
= (a^{1/2 }b^{1/2})^{-1}
By using the negative exponent a^{-x }= 1/a^{x}
= 1/(a^{1/2 }b^{1/2})
So, the answer is 1/(a^{1/2 }b^{1/2}).
Example 4 :
(x^{0}y^{1/3})^{3/2 }x^{0}
Solution :
= (x^{0}y^{1/3})^{3/2 }x^{0}
By using the zero exponent a^{0 }= 1, we get
= (1y^{1/3})^{3/2 }1
= (y^{1/3})^{3/2}
By using the power rule (a^{m})^{n }= a^{mn}, we get
= y^{(1/3 ⋅ 3/2)}
= y^{3/6}
= y^{1/2}
So, the answer is y^{1/2}.
Example 5 :
(2x^{1/}^{2}y^{1/3})/(2x^{4/3}y^{-7/4})
Solution :
= (2x^{1/}^{2}y^{1/3})/(2x^{4/3}y^{-7/4})
By writing each term separately, we get
= x^{1/}^{2 }⋅ y^{1/3 }⋅ 1/x^{4/3 }⋅ 1/y^{-7/4}
By combining the like terms, we get
= x^{1/}^{2 }⋅ 1/x^{4/3 }⋅ y^{1/3 }⋅ 1/y^{-7/4 }
By using the quotient rule a^{m}/a^{n} = a^{m-n}, we get
= x^{1/2-4/3 }⋅ y^{1/3+7/4}
= x^{-5/6}⋅ y^{25/12}
By using the negative exponent a^{-x }= 1/a^{x}
= y^{25/12 }⋅ 1/x^{5/6}
= y^{25/12}/x^{5/6}
So, the answer is y^{25/12}/x^{5/6}.
Example 6 :
u^{-}^{5/4}v^{2 }⋅ (u^{3/2})^{-3/2}
Solution :
= u^{-}^{5/4}v^{2 }⋅ (u^{3/2})^{-3/2}
= u^{-}^{5/4 }⋅ v^{2 }⋅ (u^{3/2})^{-3/2}
By using the power rule (a^{m})^{n }= a^{mn}, we get
= u^{-}^{5/4 }⋅ v^{2 }⋅ u^{(}^{3/2 }^{⋅ (}^{-3/2))}
= u^{-}^{5/4 }⋅ v^{2 }⋅ u^{-9/4}
By combining the like terms, we get
= u^{-}^{5/4 }⋅ u^{-9/4 }⋅ v^{2}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= u^{-5/4-9/4}^{ }⋅ v^{2}
= u^{-14/4 }⋅ v^{2}
= u^{-7/2 }⋅ v^{2}
By using the negative exponent a^{-x }= 1/a^{x}
= 1/u^{7/2 }⋅ v^{2}
= v^{2}/u^{7/2}
So, the answer is v^{2}/u^{7/2}.
Example 7 :
uv ⋅ u ⋅ (v^{3/2})^{3}
Solution :
= uv ⋅ u ⋅ (v^{3/2})^{3}
= u ⋅ v ⋅ u ⋅ (v^{3/2})^{3}
By using the power rule (a^{m})^{n }= a^{mn}, we get
= u ⋅ v ⋅ u ⋅ v^{9/2}
By combining the like terms, we get
= u ⋅ u ⋅ v ⋅ v^{9/2}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= u^{1+1 }⋅ v^{1+9/2}
= u^{2 }⋅ v^{11/2}
So, the answer is u^{2 }⋅ v^{11/2}.
Example 8 :
(2x^{-2}y^{5/3})/x^{-5/4}y^{-5/3 }⋅ xy^{1/2}
Solution :
= (2x^{-2}y^{5/3})/x^{-5/4}y^{-5/3 }⋅ xy^{1/2}
By writing each term separately, we get
= 2 ⋅ x^{-2}y^{5/3 }⋅ 1/x^{-5/4}^{ }⋅ 1/y^{-5/3 }⋅ 1/xy^{1/2}
By using the negative exponent 1/a^{-}^{x} = a^{x}, we get
= 2 ⋅ x^{-2}y^{5/3}^{ }⋅ x^{5/4 }⋅ y^{5/3 }⋅ 1/xy^{1/2}
By using the negative exponent 1/a^{x }= a^{-x}, we get
= 2 ⋅ x^{-2}y^{5/3}^{ }⋅ x^{5/4 }⋅ y^{5/3 }⋅ x^{-1 }⋅ y^{-1/2}
By combining the like terms, we get
= 2 ⋅ x^{-2}^{ }⋅ x^{5/4 }⋅ x^{-1 }⋅ y^{5/3}^{ }⋅ y^{5/3 }⋅ y^{-1/2}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= 2 ⋅ x^{-2+5/4}^{ }⋅ x^{-1}^{ }⋅ y^{5/3+5/3 }⋅ y^{-1/2}
= 2 ⋅ x^{-3/4 }⋅ x^{-1 }⋅ y^{10/3 }⋅ y^{-1/2}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= 2 ⋅ x^{-3/4-1 }⋅ y^{1}^{0/3-1/2}
= 2 ⋅ x^{-7/4 }⋅ y^{17/6}
By using the negative exponent a^{-x }= 1/a^{x}
= 2 ⋅ y^{17/6 }⋅ 1/x^{7/4}
= (2y^{17/6})/x^{7/4}
So, the answer is (2y^{17/6})/x^{7/4}.
Example 9 :
(a^{3/4}b^{-1 }⋅ b^{7/4})/3b^{-1}
Solution :
= (a^{3/4}b^{-1 }⋅ b^{7/4})/3b^{-1}
By writing each term separately, we get
= a^{3/4}^{ }⋅ b^{-1 }⋅ b^{7/4}^{ }⋅ 1/3^{ }⋅ b^{-1}
By using the negative exponent 1/a^{-}^{x} = a^{x}, we get
= a^{3/4}^{ }⋅ b^{-1 }⋅ b^{7/4}^{ }⋅ 1/3 ⋅ b
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= a^{3/4}^{ }⋅ b^{-1+7/4 }⋅ 1/3 ⋅ b
= a^{3/4}^{ }⋅ b^{3/4 }⋅ b ⋅ 1/3
= a^{3/4}^{ }⋅ b^{3/4+1 }⋅ 1/3
= a^{3/4}^{ }⋅ b^{7/4 }⋅ 1/3
= (a^{3/4}b^{7/4})/3
So, the answer is (a^{3/4}b^{7/4})/3.
Example 10 :
(3y^{-5/4})/y^{-1}^{ }⋅ 2y^{-1/3}
Solution :
= (3y^{-5/4})/y^{-1}^{ }⋅ 2y^{-1/3}
By writing each term separately, we get
= 3 ⋅ y^{-5/4 }⋅ 1/y^{-1 }⋅ 1/2 ⋅ 1/y^{-1/3}
By using the negative exponent 1/a^{-}^{x} = a^{x}, we get
= 3 ⋅ y^{-5/4 }⋅ y^{1 }⋅ 1/2 ⋅ y^{1/3}
By using the product rule a^{m }⋅ a^{n }= a^{m+n}, we get
= 3 ⋅ y^{-5/4+1 }⋅ 1/2 ⋅ y^{1/3}
= 3 ⋅ y^{-1/4 }⋅ y^{1/3 }⋅ 1/2
= 3 ⋅ y^{-1/4+1/3 }⋅ 1/2
= 3 ⋅ y^{1/12 }⋅ 1/2
= (3y^{1/12})/2
So, the answer is (3y^{1/12})/2.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM