Question

Simplify: \[{\left( {64} \right)^{\dfrac{1}{2}}}\].

Answer 1

Hint:
Here, we need to find the value of the given expression. First, we will rewrite the number 64 as a product of its factors. Then, we will use rules of exponents and simplify the expression to find the value of the expression \[{\left( {64} \right)^{\dfrac{1}{2}}}\].

Complete step by step solution:
We will rewrite the number 64 as a product of its factors.
A prime number is a number which is divisible only by 2 natural numbers, that is by 1 and by the number itself.
For example: 5 is a prime number because it is divisible by only 2 natural numbers, 1 and 5.
We will check the divisibility of the number 64 by prime numbers.
First, let us check the divisibility by 2.
We know that every even number is divisible by 2. This means that any number that has any of the digits 0, 2, 4, 6, 8 in the units place is divisible by 2.
The number 64 has 4 in the units place.
Therefore, 64 is divisible by 2.
Dividing 64 by 2, we get
\[ \Rightarrow \dfrac{{64}}{2} = 32\]
Now, we will check the divisibility of 32 by 2.
The number 32 has 2 in the units place.
Therefore, 32 is divisible by 2.
Dividing 32 by 2, we get
\[ \Rightarrow \dfrac{{32}}{2} = 16\]
Next, we will check the divisibility of 16 by 2.
The number 16 has 6 in the units place.
Therefore, 16 is divisible by 2.
Dividing 16 by 2, we get
\[ \Rightarrow \dfrac{{16}}{2} = 8\]
Thus, we can write 64 as
\[ \Rightarrow 64 = 2 \times 2 \times 2 \times 8\]
We know that \[8 = 4 \times 2 = 2 \times 2 \times 2\].
Thus, we get
\[ \Rightarrow 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]
Rewriting the expression, we get
\[ \Rightarrow 64 = {2^1} \times {2^1} \times {2^1} \times {2^1} \times {2^1} \times {2^1}\]
We know that if two or more numbers with same base and different exponents are multiplied, the product can be written as \[{a^b} \times {a^c} \times {a^d} = {a^{b + c + d}}\].
Therefore, rewriting the expression using the rule of exponents, we get
\[\begin{array}{l} \Rightarrow 64 = {2^{1 + 1 + 1 + 1 + 1 + 1}}\\ \Rightarrow 64 = {2^6}\end{array}\]
Now, we will simplify the given expression.
Substituting \[64 = {2^6}\] in the expression \[{\left( {64} \right)^{\dfrac{1}{2}}}\], we get
\[ \Rightarrow {\left( {64} \right)^{\dfrac{1}{2}}} = {\left( {{2^6}} \right)^{\dfrac{1}{2}}}\]
If a number raised to an exponent is again raised to an exponent, the new power is the product of the exponents. This can be written as \[{\left( {{a^b}} \right)^c} = {a^{b \times c}}\].
Therefore, the equation becomes
\[ \Rightarrow {\left( {64} \right)^{\dfrac{1}{2}}} = {\left( 2 \right)^{6 \times \dfrac{1}{2}}}\]
Multiplying the exponents, we get
\[ \Rightarrow {\left( {64} \right)^{\dfrac{1}{2}}} = {2^3}\]
Thus, we get
\[\begin{array}{l} \Rightarrow {\left( {64} \right)^{\dfrac{1}{2}}} = 2 \times 2 \times 2\\ \Rightarrow {\left( {64} \right)^{\dfrac{1}{2}}} = 8\end{array}\]

\[\therefore \] We get the value of the expression \[{\left( {64} \right)^{\dfrac{1}{2}}}\] as 8.

Note:
We can also solve this directly using the multiplication table of 8.
We know that 64 is the product of 8 and 8.
Thus, we have
\[\begin{array}{l} \Rightarrow 64 = 8 \times 8\\ \Rightarrow 64 = {8^1} \times {8^1}\end{array}\]
Thus, we get
\[ \Rightarrow 64 = {8^2}\]
Taking the square root on both the sides, we get
\[\begin{array}{l} \Rightarrow \sqrt {64} = \sqrt {{8^2}} \\ \Rightarrow \sqrt {64} = 8\end{array}\]
The square root of a number \[x\] can be denoted by either \[\sqrt x \] or \[{x^{\dfrac{1}{2}}}\].
Therefore, we get
\[{\left( {64} \right)^{\dfrac{1}{2}}} = 8\]
\[\therefore \] We get the value of the expression \[{\left( {64} \right)^{\dfrac{1}{2}}}\] as 8.