
How do you simplify the given division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$?
Answer
562.5k+ views
Hint: We start solving the problem by equating the given $\dfrac{{{8}^{2}}}{{{8}^{2}}}$ to a variable. We then make use of the fact that $8={{2}^{3}}$ to proceed through the problem. We then make use of the law of exponents ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to proceed further through the problem. We then make use of the law of exponents $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, for $m\ge n$ to proceed further through the problem. We then make use of the law of exponents ${{a}^{0}}=1$ to get the required answer.
Complete step by step answer:
According to the problem, we are asked to simplify the division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$.
Let us assume $d=\dfrac{{{8}^{2}}}{{{8}^{2}}}$ ---(1).
We know that $8={{2}^{3}}$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{{{\left( {{2}^{3}} \right)}^{2}}}{{{\left( {{2}^{3}} \right)}^{2}}}$ ---(2).
From the law of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this result in equation (2).
$\Rightarrow d=\dfrac{{{2}^{3\times 2}}}{{{2}^{3\times 2}}}$.
$\Rightarrow d=\dfrac{{{2}^{6}}}{{{2}^{6}}}$ ---(3).
From the law of exponents, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, for $m\ge n$. Let us use this result in equation (3).
$\Rightarrow d={{2}^{6-6}}$.
$\Rightarrow d={{2}^{0}}$ ---(4).
From the law of exponents, we know that ${{a}^{0}}=1$, for $a\ne 0$. Let us use this result in equation (4).
$\Rightarrow d=1$.
So, we have found the value of the given division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$ as 1.
$\therefore $ The result of the given division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$ is 1.
Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not take ${{8}^{2}}={{2}^{{{3}^{2}}}}$ instead of ${{8}^{2}}={{\left( {{2}^{3}} \right)}^{2}}$ which is the common mistakes done by students. We can also solve the problem by making use of the fact that ${{8}^{2}}=64$. Similarly, we can expect problems to find the result of the given division $\dfrac{{{8}^{3}}}{{{16}^{2}}}$.
Complete step by step answer:
According to the problem, we are asked to simplify the division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$.
Let us assume $d=\dfrac{{{8}^{2}}}{{{8}^{2}}}$ ---(1).
We know that $8={{2}^{3}}$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{{{\left( {{2}^{3}} \right)}^{2}}}{{{\left( {{2}^{3}} \right)}^{2}}}$ ---(2).
From the law of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this result in equation (2).
$\Rightarrow d=\dfrac{{{2}^{3\times 2}}}{{{2}^{3\times 2}}}$.
$\Rightarrow d=\dfrac{{{2}^{6}}}{{{2}^{6}}}$ ---(3).
From the law of exponents, we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, for $m\ge n$. Let us use this result in equation (3).
$\Rightarrow d={{2}^{6-6}}$.
$\Rightarrow d={{2}^{0}}$ ---(4).
From the law of exponents, we know that ${{a}^{0}}=1$, for $a\ne 0$. Let us use this result in equation (4).
$\Rightarrow d=1$.
So, we have found the value of the given division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$ as 1.
$\therefore $ The result of the given division $\dfrac{{{8}^{2}}}{{{8}^{2}}}$ is 1.
Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not take ${{8}^{2}}={{2}^{{{3}^{2}}}}$ instead of ${{8}^{2}}={{\left( {{2}^{3}} \right)}^{2}}$ which is the common mistakes done by students. We can also solve the problem by making use of the fact that ${{8}^{2}}=64$. Similarly, we can expect problems to find the result of the given division $\dfrac{{{8}^{3}}}{{{16}^{2}}}$.
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