How do you evaluate ${{\left( {{16}^{\dfrac{1}{2}}} \right)}^{3}}$?
Answer
615k+ views
Hint: Assume the given expression as ‘E’. Now, use the prime factorization method and write the base of the expression, which is 16, into its exponential form. In the next step, apply the formula of exponents and powers as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] to simplify the value of the expression.
Complete step-by-step answer:
Here, we have been provided with the expression ${{\left( {{16}^{\dfrac{1}{2}}} \right)}^{3}}$ and we are asked to evaluate it.
Now, let us assume the value of the given expression as E. So, we have,
$\Rightarrow E={{\left( {{16}^{\dfrac{1}{2}}} \right)}^{3}}$
Now, we need to convert the base of this expression, i.e. 16, into the exponential form by using the prime factorization method, so writing the base 16 as the product of its primes we get the expression as:
\[\Rightarrow E={{\left( {{\left( 2\times 2\times 2\times 2 \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
Converting the above expression into the exponential form we get,
\[\Rightarrow E={{\left( {{\left( {{2}^{4}} \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
Using the formula of exponents given as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
& \Rightarrow E={{\left( {{2}^{4\times \dfrac{1}{2}}} \right)}^{3}} \\
& \Rightarrow E={{\left( {{2}^{2}} \right)}^{3}} \\
& \Rightarrow E={{4}^{3}} \\
& \Rightarrow E=4\times 4\times 4 \\
& \Rightarrow E=64 \\
\end{align}\]
Hence, the above expression represents the simplified form of the given exponential expression.
Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}},{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] as they are used everywhere. As you may know that the exponent $\dfrac{1}{2}$ represents the square root of any base, so you can directly say that the square root of 16 will be 4 and hence you will not be required to use the prime factorization. The method of prime factorization is used when we have large numbers as the base.
Complete step-by-step answer:
Here, we have been provided with the expression ${{\left( {{16}^{\dfrac{1}{2}}} \right)}^{3}}$ and we are asked to evaluate it.
Now, let us assume the value of the given expression as E. So, we have,
$\Rightarrow E={{\left( {{16}^{\dfrac{1}{2}}} \right)}^{3}}$
Now, we need to convert the base of this expression, i.e. 16, into the exponential form by using the prime factorization method, so writing the base 16 as the product of its primes we get the expression as:
\[\Rightarrow E={{\left( {{\left( 2\times 2\times 2\times 2 \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
Converting the above expression into the exponential form we get,
\[\Rightarrow E={{\left( {{\left( {{2}^{4}} \right)}^{\dfrac{1}{2}}} \right)}^{3}}\]
Using the formula of exponents given as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we get,
\[\begin{align}
& \Rightarrow E={{\left( {{2}^{4\times \dfrac{1}{2}}} \right)}^{3}} \\
& \Rightarrow E={{\left( {{2}^{2}} \right)}^{3}} \\
& \Rightarrow E={{4}^{3}} \\
& \Rightarrow E=4\times 4\times 4 \\
& \Rightarrow E=64 \\
\end{align}\]
Hence, the above expression represents the simplified form of the given exponential expression.
Note: One may note that here we have used some basic formulas of the topic ‘exponents and powers’ to solve the question. You must remember some basic formulas such as: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}},{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] as they are used everywhere. As you may know that the exponent $\dfrac{1}{2}$ represents the square root of any base, so you can directly say that the square root of 16 will be 4 and hence you will not be required to use the prime factorization. The method of prime factorization is used when we have large numbers as the base.
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