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Hint: The given question is related to power and indices. Try to recall the formulae related to mathematical operations of numbers in their exponent form.

Complete step-by-step answer:

Before proceeding with the solution, we must know the formulae used to solve the problem.

A) $\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}$

B) $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$

C) $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$

D) ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$

Now, we will use these formulae to solve the question. First, we will express the numbers in their exponent form. We know we can write $\sqrt[n]{a}$ as ${{a}^{\dfrac{1}{n}}}$ . So, we can write $\sqrt[3]{24}$ as ${{24}^{\dfrac{1}{3}}}$ , $\sqrt[4]{2}$ as ${{2}^{\dfrac{1}{4}}}$ and $\sqrt[3]{3}$ as ${{3}^{\dfrac{1}{3}}}$ . So, we can write $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ as ${{24}^{\dfrac{1}{3}}}\div \left( {{2}^{\dfrac{1}{4}}}\times {{3}^{\dfrac{1}{3}}} \right)$.

We can see that the indices of $24$ and $3$ are the same. Also, we know that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$ . So, ${{24}^{\dfrac{1}{3}}}\div {{3}^{\dfrac{1}{3}}}={{\left( \dfrac{24}{3} \right)}^{\dfrac{1}{3}}}={{8}^{\dfrac{1}{3}}}$ . So, $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ becomes ${{8}^{\dfrac{1}{3}}}\div {{2}^{\dfrac{1}{4}}}$ .

Now, we know that we can write $8$ as ${{2}^{3}}$ . So, we can write ${{8}^{\dfrac{1}{3}}}$ as ${{\left( {{2}^{3}} \right)}^{\dfrac{1}{3}}}=2$ . So, ${{8}^{\dfrac{1}{3}}}\div {{2}^{\dfrac{1}{4}}}$ becomes $2\div {{2}^{\dfrac{1}{4}}}$ . Now, in $2\div {{2}^{\dfrac{1}{4}}}$ , the bases are the same, i.e. $2$ . Also, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. So, $2\div {{2}^{\dfrac{1}{4}}}={{2}^{1-\dfrac{1}{4}}}={{2}^{\dfrac{3}{4}}}$ .

Now, we know that the value of a fraction does not change on multiplying and dividing it by the same number. So, we can write ${{2}^{\dfrac{3}{4}}}$ as ${{2}^{\dfrac{3}{4}\times \dfrac{3}{3}}}={{2}^{\dfrac{9}{12}}}$. We know, ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$ . So, we can write \[{{2}^{\dfrac{9}{12}}}\] as \[{{\left( {{2}^{9}} \right)}^{\dfrac{1}{12}}}\].

Now, we know \[{{a}^{m}}=a\times a\times a\times a....m\,times\]. So, ${{2}^{9}}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=512$ . So, \[{{\left( {{2}^{9}} \right)}^{\dfrac{1}{12}}}={{512}^{\dfrac{1}{12}}}\]. We know we can write ${{a}^{\dfrac{1}{n}}}$ as $\sqrt[n]{a}$ . So, we can write \[{{512}^{\dfrac{1}{12}}}\] as $\sqrt[12]{512}$ . So, the value of $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ is equal to $\sqrt[12]{512}$ .

Hence, option A. is the right answer.

Note: ${{\left( {{a}^{m}} \right)}^{n}}$ and ${{a}^{{{m}^{n}}}}$ are not same. In ${{\left( {{a}^{m}} \right)}^{n}}$ , the value ${{a}^{m}}$ is raised to the power $n$ , whereas in ${{a}^{{{m}^{n}}}}$, only the exponent $m$ is raised to the power $n$. Students generally get confused and treat both as the same and end up getting a wrong answer.

Complete step-by-step answer:

Before proceeding with the solution, we must know the formulae used to solve the problem.

A) $\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}$

B) $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$

C) $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$

D) ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$

Now, we will use these formulae to solve the question. First, we will express the numbers in their exponent form. We know we can write $\sqrt[n]{a}$ as ${{a}^{\dfrac{1}{n}}}$ . So, we can write $\sqrt[3]{24}$ as ${{24}^{\dfrac{1}{3}}}$ , $\sqrt[4]{2}$ as ${{2}^{\dfrac{1}{4}}}$ and $\sqrt[3]{3}$ as ${{3}^{\dfrac{1}{3}}}$ . So, we can write $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ as ${{24}^{\dfrac{1}{3}}}\div \left( {{2}^{\dfrac{1}{4}}}\times {{3}^{\dfrac{1}{3}}} \right)$.

We can see that the indices of $24$ and $3$ are the same. Also, we know that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$ . So, ${{24}^{\dfrac{1}{3}}}\div {{3}^{\dfrac{1}{3}}}={{\left( \dfrac{24}{3} \right)}^{\dfrac{1}{3}}}={{8}^{\dfrac{1}{3}}}$ . So, $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ becomes ${{8}^{\dfrac{1}{3}}}\div {{2}^{\dfrac{1}{4}}}$ .

Now, we know that we can write $8$ as ${{2}^{3}}$ . So, we can write ${{8}^{\dfrac{1}{3}}}$ as ${{\left( {{2}^{3}} \right)}^{\dfrac{1}{3}}}=2$ . So, ${{8}^{\dfrac{1}{3}}}\div {{2}^{\dfrac{1}{4}}}$ becomes $2\div {{2}^{\dfrac{1}{4}}}$ . Now, in $2\div {{2}^{\dfrac{1}{4}}}$ , the bases are the same, i.e. $2$ . Also, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. So, $2\div {{2}^{\dfrac{1}{4}}}={{2}^{1-\dfrac{1}{4}}}={{2}^{\dfrac{3}{4}}}$ .

Now, we know that the value of a fraction does not change on multiplying and dividing it by the same number. So, we can write ${{2}^{\dfrac{3}{4}}}$ as ${{2}^{\dfrac{3}{4}\times \dfrac{3}{3}}}={{2}^{\dfrac{9}{12}}}$. We know, ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$ . So, we can write \[{{2}^{\dfrac{9}{12}}}\] as \[{{\left( {{2}^{9}} \right)}^{\dfrac{1}{12}}}\].

Now, we know \[{{a}^{m}}=a\times a\times a\times a....m\,times\]. So, ${{2}^{9}}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=512$ . So, \[{{\left( {{2}^{9}} \right)}^{\dfrac{1}{12}}}={{512}^{\dfrac{1}{12}}}\]. We know we can write ${{a}^{\dfrac{1}{n}}}$ as $\sqrt[n]{a}$ . So, we can write \[{{512}^{\dfrac{1}{12}}}\] as $\sqrt[12]{512}$ . So, the value of $\sqrt[3]{24}\div \left( \sqrt[4]{2}.\sqrt[3]{3} \right)$ is equal to $\sqrt[12]{512}$ .

Hence, option A. is the right answer.

Note: ${{\left( {{a}^{m}} \right)}^{n}}$ and ${{a}^{{{m}^{n}}}}$ are not same. In ${{\left( {{a}^{m}} \right)}^{n}}$ , the value ${{a}^{m}}$ is raised to the power $n$ , whereas in ${{a}^{{{m}^{n}}}}$, only the exponent $m$ is raised to the power $n$. Students generally get confused and treat both as the same and end up getting a wrong answer.

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