How do you simplify $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}$ ?
Answer
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Hint: For answering this question we have been asked to simplify the given expression $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}$ . For simplifying the expression we will use the formulae given as $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ from the concepts of exponents. We know that the number eighty one can be expressed nine raised to the power of two we will use this in our solution.
Complete step by step solution:
Here from the question we have been asked to simplify the expression $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}$ .
From the basic concepts of exponents we know that we have a formula given as $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ if we observe given expression is in similar form of this formula so we will use this formula to simplify the expression.
After using we will have $\Rightarrow \dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}={{81}^{\dfrac{3}{4}-\dfrac{1}{4}}}$ .
Now we will perform subtraction operation between the numbers in the exponents after performing that we will have $\Rightarrow {{81}^{\dfrac{3}{4}-\dfrac{1}{4}}}={{81}^{\dfrac{2}{4}}}$ .
Now for further simplifying the expression we will reduce the fraction in the exponent after doing this we will have $\Rightarrow {{81}^{\dfrac{2}{4}}}={{81}^{\dfrac{1}{2}}}$ .
Here we can apply the concept that states that the square root of any number is equivalent to the number raised to the power half.
By applying this in the given expression we will have ${{81}^{\dfrac{1}{2}}}=\sqrt{81}$ .
As we know that the number eighty one can be written as the square of nine. We will simply write the expression as
$\begin{align}
& \Rightarrow \sqrt{81}=\sqrt{{{9}^{2}}} \\
& \Rightarrow 9 \\
\end{align}$
Therefore we can conclude that the simplified form of the given number is $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}=9$ .
Note: While answering questions of this type we should be sure with our concepts and calculations. Similarly we can simplify any expression using the exponent formula some of the exponent formulae are ${{a}^{x}}{{a}^{y}}={{a}^{x+y}}$ , ${{a}^{x-y}}=\dfrac{{{a}^{x}}}{{{a}^{y}}}$ , ${{a}^{-x}}=\left( \dfrac{1}{{{a}^{x}}} \right)$ , ${{a}^{0}}=1$ and many more.
Complete step by step solution:
Here from the question we have been asked to simplify the expression $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}$ .
From the basic concepts of exponents we know that we have a formula given as $\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}$ if we observe given expression is in similar form of this formula so we will use this formula to simplify the expression.
After using we will have $\Rightarrow \dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}={{81}^{\dfrac{3}{4}-\dfrac{1}{4}}}$ .
Now we will perform subtraction operation between the numbers in the exponents after performing that we will have $\Rightarrow {{81}^{\dfrac{3}{4}-\dfrac{1}{4}}}={{81}^{\dfrac{2}{4}}}$ .
Now for further simplifying the expression we will reduce the fraction in the exponent after doing this we will have $\Rightarrow {{81}^{\dfrac{2}{4}}}={{81}^{\dfrac{1}{2}}}$ .
Here we can apply the concept that states that the square root of any number is equivalent to the number raised to the power half.
By applying this in the given expression we will have ${{81}^{\dfrac{1}{2}}}=\sqrt{81}$ .
As we know that the number eighty one can be written as the square of nine. We will simply write the expression as
$\begin{align}
& \Rightarrow \sqrt{81}=\sqrt{{{9}^{2}}} \\
& \Rightarrow 9 \\
\end{align}$
Therefore we can conclude that the simplified form of the given number is $\dfrac{{{81}^{\dfrac{3}{4}}}}{{{81}^{\dfrac{1}{4}}}}=9$ .
Note: While answering questions of this type we should be sure with our concepts and calculations. Similarly we can simplify any expression using the exponent formula some of the exponent formulae are ${{a}^{x}}{{a}^{y}}={{a}^{x+y}}$ , ${{a}^{x-y}}=\dfrac{{{a}^{x}}}{{{a}^{y}}}$ , ${{a}^{-x}}=\left( \dfrac{1}{{{a}^{x}}} \right)$ , ${{a}^{0}}=1$ and many more.
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