
How do you express \[{{t}^{-\dfrac{2}{7}}}\] in radical form?
Answer
541.8k+ views
Hint: In order to find the solution to the given question, that is to express \[{{t}^{-\dfrac{2}{7}}}\] in radical form, apply the rule for exponents that is \[{{x}^{a\times b}}={{\left( {{x}^{a}} \right)}^{b}}\] then to convert the given expression into radical form, apply following formulas: \[{{x}^{\dfrac{1}{n}}}=\sqrt[n]{x}\] and \[{{x}^{-a}}=\dfrac{1}{{{x}^{a}}}\]. Here expressing in radical form means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.
Complete step-by-step answer:
According to the question, given expression in the question is as follows:
\[{{t}^{-\dfrac{2}{7}}}\]
We can rewrite the above expression as:
\[\Rightarrow {{t}^{-2\times \dfrac{1}{7}}}\]
Now, apply the rule for exponents that is \[{{x}^{a\times b}}={{\left( {{x}^{a}} \right)}^{b}}\] to rewrite the expression again as:
\[\Rightarrow {{t}^{-2\times \dfrac{1}{7}}}={{\left( {{t}^{-2}} \right)}^{\dfrac{1}{7}}}\]
In order to convert the above expression into radical form. Here expressing in radical form means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.
So first apply the formula: \[{{x}^{\dfrac{1}{n}}}=\sqrt[n]{x}\] in the above expression, we get:
\[\Rightarrow {{\left( {{t}^{-2}} \right)}^{\dfrac{1}{7}}}=\sqrt[7]{{{t}^{-2}}}\]
Secondly, in order to have no negative exponents we can use this rule of exponents to eliminate the negative exponent that is \[{{x}^{-a}}=\dfrac{1}{{{x}^{a}}}\] in the above expression, we get:
\[\Rightarrow \sqrt[7]{{{t}^{-2}}}=\sqrt[7]{\dfrac{1}{{{t}^{2}}}}\]
We know that the nth root of 1 is always 1, so by applying this concept in the above expression, we get:
\[\Rightarrow \sqrt[7]{\dfrac{1}{{{t}^{2}}}}=\dfrac{1}{\sqrt[7]{{{t}^{2}}}}\]
Therefore, radical form of the given expression \[{{t}^{-\dfrac{2}{7}}}\] is \[\dfrac{1}{\sqrt[7]{{{t}^{2}}}}\].
Note: Students generally face problems while understanding the concept of radical form. It’s important to know that expressing in radical form means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.
Complete step-by-step answer:
According to the question, given expression in the question is as follows:
\[{{t}^{-\dfrac{2}{7}}}\]
We can rewrite the above expression as:
\[\Rightarrow {{t}^{-2\times \dfrac{1}{7}}}\]
Now, apply the rule for exponents that is \[{{x}^{a\times b}}={{\left( {{x}^{a}} \right)}^{b}}\] to rewrite the expression again as:
\[\Rightarrow {{t}^{-2\times \dfrac{1}{7}}}={{\left( {{t}^{-2}} \right)}^{\dfrac{1}{7}}}\]
In order to convert the above expression into radical form. Here expressing in radical form means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.
So first apply the formula: \[{{x}^{\dfrac{1}{n}}}=\sqrt[n]{x}\] in the above expression, we get:
\[\Rightarrow {{\left( {{t}^{-2}} \right)}^{\dfrac{1}{7}}}=\sqrt[7]{{{t}^{-2}}}\]
Secondly, in order to have no negative exponents we can use this rule of exponents to eliminate the negative exponent that is \[{{x}^{-a}}=\dfrac{1}{{{x}^{a}}}\] in the above expression, we get:
\[\Rightarrow \sqrt[7]{{{t}^{-2}}}=\sqrt[7]{\dfrac{1}{{{t}^{2}}}}\]
We know that the nth root of 1 is always 1, so by applying this concept in the above expression, we get:
\[\Rightarrow \sqrt[7]{\dfrac{1}{{{t}^{2}}}}=\dfrac{1}{\sqrt[7]{{{t}^{2}}}}\]
Therefore, radical form of the given expression \[{{t}^{-\dfrac{2}{7}}}\] is \[\dfrac{1}{\sqrt[7]{{{t}^{2}}}}\].
Note: Students generally face problems while understanding the concept of radical form. It’s important to know that expressing in radical form means simplifying a radical so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any radicals in the denominator of a fraction.
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