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If \[\sqrt {{m^4}{n^4}} \times \sqrt[6]{{{m^2}{n^2}}} \times \sqrt[3]{{{m^2}{n^2}}} = {\left( {mn} \right)^k}\], then find the value of k?
(a) 6
(b) 3
(c) 2
(d) 1

Answer
VerifiedVerified
149.4k+ views
Hint- First, We should convert the under root into its numerical value and try to bring in power of mn then add up all the power of mn which is equal to k.

Replacing the under-roots with their values, we get
$
  {\left( {{m^4}{n^4}} \right)^{\dfrac{1}{2}}} \times {\left( {{m^2}{n^2}} \right)^{\dfrac{1}{6}}} \times {\left( {{m^2}{n^2}} \right)^{\dfrac{1}{3}}} = {\left( {mn} \right)^k} \\
    \\
$
Now, using the properties of exponents
     $
   \Rightarrow {\left( {{m^4}{n^4}} \right)^{\dfrac{1}{2}}} \times {\left( {{m^2}{n^2}} \right)^{\dfrac{1}{6}}} \times {\left( {{m^2}{n^2}} \right)^{\dfrac{1}{3}}} = {\left( {mn} \right)^k} \\
   \Rightarrow \because {\left( {{a^x}} \right)^y} = {a^{xy}} \\
   \Rightarrow {\left( {mn} \right)^{\dfrac{4}{2}}} \times {\left( {mn} \right)^{\dfrac{2}{6}}} \times {\left( {mn} \right)^{\dfrac{2}{3}}} = {\left( {mn} \right)^k} \\
   \Rightarrow \because {a^x}{a^y} = {a^{x + y}} \\
   \Rightarrow {\left( {mn} \right)^2} \times {\left( {mn} \right)^{\dfrac{1}{3}}} \times {\left( {mn} \right)^{\dfrac{2}{3}}} = {\left( {mn} \right)^k} \\
   \Rightarrow {\left( {mn} \right)^{2 + \dfrac{1}{3} + \dfrac{2}{3}}} = {\left( {mn} \right)^k} \\
   \Rightarrow {\left( {mn} \right)^3} = {\left( {mn} \right)^k} \\
   \Rightarrow \therefore k = 3 \\
$
∴the correct option is B.

Note- In this question we used the property saying when the bases are same powers get added Some more properties of exponents other than the two used in the above questions:
1. \[{\left( {ab} \right)^m} = {a^{^m}}{b^m}\]
2. \[{a^{ - n}} = {a^{\dfrac{1}{n}}}\]
3. \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\]
4. \[{\left( {\dfrac{a}{b}} \right)^{ - n}} = {\left( {\dfrac{b}{a}} \right)^n}\]
5. \[{a^0} = 1\]