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# Simplify and express the result in its simplest form: $\sqrt{24}\div \left( \sqrt{2}.\sqrt{3} \right)$ .A) $\sqrt{512}$B) $\sqrt{502}$C) $\sqrt{256}$D) $\sqrt{512}$ Verified
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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.

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 $\sqrt{24}$ as ${{24}^{\dfrac{1}{3}}}$ , $\sqrt{2}$ as ${{2}^{\dfrac{1}{4}}}$ and $\sqrt{3}$ as ${{3}^{\dfrac{1}{3}}}$ . So, we can write $\sqrt{24}\div \left( \sqrt{2}.\sqrt{3} \right)$ 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, $\sqrt{24}\div \left( \sqrt{2}.\sqrt{3} \right)$ 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 $\sqrt{512}$ . So, the value of $\sqrt{24}\div \left( \sqrt{2}.\sqrt{3} \right)$ is equal to $\sqrt{512}$ .
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.
Last updated date: 02nd Oct 2023
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