How do you simplify \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\]?
Answer
624.6k+ views
Hint: This question belongs to the topic of exponents of the chapter algebra. In solving this question, we will first find the square root of the term \[\dfrac{16}{25}\] because in the denominator of the power of this term it is given as 2. After finding the square root of that term, we will find the cube of the new term. Then, we will get the answer.
Complete step by step answer:
Let us solve this question.
In this question, we have to simplify the term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] or we can say we have to solve this term and remove the powers of the term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\].
The term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] also can be written as
\[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{16}{25} \right)}^{3\times \dfrac{1}{2}}}\]
The above equation also can be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{16}{25} \right)}^{\dfrac{1}{2}\times 3}}\]
As we know that the term \[{{a}^{m\times n}}\] can be written as \[{{\left( {{a}^{m}} \right)}^{n}}\].
So, we can write the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( {{\left( \dfrac{16}{25} \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
As we know that the term \[{{a}^{\dfrac{1}{n}}}\] can also be written as \[\sqrt[n]{a}\]
Or we can say that the term \[{{a}^{\dfrac{1}{2}}}\] can also be written as \[\sqrt[2]{a}\]
So, the above equation also can be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \sqrt[2]{\dfrac{16}{25}} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \sqrt{\dfrac{16}{25}} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{\sqrt{16}}{\sqrt{25}} \right)}^{3}}\]
As we know that the square roots of 16 and 25 are 4 and 5 respectively.
So, we can write the above the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{4}{5} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}=\dfrac{{{\left( 4 \right)}^{3}}}{{{\left( 5 \right)}^{3}}}\]
As we know that the cubes of 4 and 5 are 64 and 125 respectively.
So, we can write the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}=\dfrac{{{\left( 4 \right)}^{3}}}{{{\left( 5 \right)}^{3}}}\]
Hence, the simplified value of \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] is \[\dfrac{64}{125}\].
Note:
We should have a better knowledge in the topic of exponents to solve this type of question easily. Remember the formulas like:
\[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
And don’t forget the square roots of some specific numbers like 25 and 16. And, also don’t forget the cubes of some specific numbers like 4 and 5 to solve this type of question easily.
Complete step by step answer:
Let us solve this question.
In this question, we have to simplify the term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] or we can say we have to solve this term and remove the powers of the term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\].
The term \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] also can be written as
\[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{16}{25} \right)}^{3\times \dfrac{1}{2}}}\]
The above equation also can be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{16}{25} \right)}^{\dfrac{1}{2}\times 3}}\]
As we know that the term \[{{a}^{m\times n}}\] can be written as \[{{\left( {{a}^{m}} \right)}^{n}}\].
So, we can write the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( {{\left( \dfrac{16}{25} \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
As we know that the term \[{{a}^{\dfrac{1}{n}}}\] can also be written as \[\sqrt[n]{a}\]
Or we can say that the term \[{{a}^{\dfrac{1}{2}}}\] can also be written as \[\sqrt[2]{a}\]
So, the above equation also can be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \sqrt[2]{\dfrac{16}{25}} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \sqrt{\dfrac{16}{25}} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{\sqrt{16}}{\sqrt{25}} \right)}^{3}}\]
As we know that the square roots of 16 and 25 are 4 and 5 respectively.
So, we can write the above the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}={{\left( \dfrac{4}{5} \right)}^{3}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}=\dfrac{{{\left( 4 \right)}^{3}}}{{{\left( 5 \right)}^{3}}}\]
As we know that the cubes of 4 and 5 are 64 and 125 respectively.
So, we can write the above equation as
\[\Rightarrow {{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}=\dfrac{{{\left( 4 \right)}^{3}}}{{{\left( 5 \right)}^{3}}}\]
Hence, the simplified value of \[{{\left( \dfrac{16}{25} \right)}^{\dfrac{3}{2}}}\] is \[\dfrac{64}{125}\].
Note:
We should have a better knowledge in the topic of exponents to solve this type of question easily. Remember the formulas like:
\[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
And don’t forget the square roots of some specific numbers like 25 and 16. And, also don’t forget the cubes of some specific numbers like 4 and 5 to solve this type of question easily.
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