Find the value of \[\sqrt[4]{{{{(64)}^{ - 2}}}}\]?
Answer
602.4k+ views
Hint: The given question is to solve by simplifying the power and root, here first the negative power on the number under root is needed to be solved, then after root value is needed to be found out. Here we know that negative power means that the number is needed to be written in reciprocal form, to remove the minus sign.
Formulae Used: To remove the minus sign from power we can reciprocate the given number:
\[ \Rightarrow {a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Complete step-by-step solution:
Here to solve the given question, first we need to solve the power, on the number:
\[
\Rightarrow \sqrt[4]{{{{(64)}^{ - 2}}}} \\
\Rightarrow \sqrt[4]{{\dfrac{1}{{{{64}^2}}}}} \\
\Rightarrow \sqrt[4]{{{{\left( {\dfrac{1}{8}} \right)}^4}}} = {\left( {\dfrac{1}{8}} \right)^{4 \times \dfrac{1}{4}}} = {\left( {\dfrac{1}{8}} \right)^1} = \dfrac{1}{8} \\
\]
Here we got the final simplified answer.
Note: The given question needs to be simplified by solving it to the shortest term, here we simplified the term to the least small answer possible. Here to check the solution to be correct, the answer can be cross checked by solving the question from solution back to question.
Formulae Used: To remove the minus sign from power we can reciprocate the given number:
\[ \Rightarrow {a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Complete step-by-step solution:
Here to solve the given question, first we need to solve the power, on the number:
\[
\Rightarrow \sqrt[4]{{{{(64)}^{ - 2}}}} \\
\Rightarrow \sqrt[4]{{\dfrac{1}{{{{64}^2}}}}} \\
\Rightarrow \sqrt[4]{{{{\left( {\dfrac{1}{8}} \right)}^4}}} = {\left( {\dfrac{1}{8}} \right)^{4 \times \dfrac{1}{4}}} = {\left( {\dfrac{1}{8}} \right)^1} = \dfrac{1}{8} \\
\]
Here we got the final simplified answer.
Note: The given question needs to be simplified by solving it to the shortest term, here we simplified the term to the least small answer possible. Here to check the solution to be correct, the answer can be cross checked by solving the question from solution back to question.
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