Question

Simplify \[{\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3}\] and find the negative of power of \[q\].

Answer 1

Hint:
Here we will first simplify the given expression using the properties of the exponential function. Then we will use the BODMAS rule to simplify it further and at last we will multiply the obtained power by negative sign to find the required value of the negative power of the given variable.

Complete step by step solution:
The given mathematical expression is \[{\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3}\].
Now, we will use the properties of the exponential function.
We know from the basic properties of exponential function that if we take the power of the exponential then the powers get multiplied.
Using the property of exponential function \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\], we get
\[{\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = {2^2}{q^{ - 3 \times 2}} \div {4^{\dfrac{1}{2}}}{q^{\dfrac{1}{2}}} \times {2^3}{q^{\dfrac{1}{6} \times 3}}\]
On further simplifying the terms, we get
\[ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 4{q^{ - 6}} \div 2{q^{\dfrac{1}{2}}} \times 8{q^{\dfrac{1}{2}}}\]
Now, we will use the BODMAS rule to solve the given mathematical expression and we will follow the same order of mathematical operations according to the BODMAS rule.
We will first divide the first two terms here.
\[ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 2{q^{ - 6 - \dfrac{1}{2}}} \times 8{q^{\dfrac{1}{2}}}\]
Now, we will simplify the power of first term further.
\[\begin{array}{l} \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 2{q^{\dfrac{{ - 12 - 1}}{2}}} \times 8{q^{\dfrac{1}{2}}}\\ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 2{q^{\dfrac{{ - 13}}{2}}} \times 8{q^{\dfrac{1}{2}}}\end{array}\]
Now, we will multiply the terms here.
\[ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 16{q^{\dfrac{{ - 13}}{2} + \dfrac{1}{2}}}\]
Now, we will simplify the power of term.
 \[ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 16{q^{\dfrac{{ - 12}}{2}}}\]
On further simplifying the term, we get
\[ \Rightarrow {\left( {2{q^{ - 3}}} \right)^2} \div {\left( {4q} \right)^{\dfrac{1}{2}}} \times {\left( {2{q^{\dfrac{1}{6}}}} \right)^3} = 16{q^{ - 6}}\]
Now, we will find the negative of the power of \[q\].
Therefore, the negative of power of \[ - 6\] is equal to 6.

Therefore, the correct answer is 6.

Note:
Here we have applied the BODMAS rule to get the value of the given mathematical operation and it stands for brackets, order, division, multiplication, addition and then subtraction. This rule explains the order of mathematical operation to solve any mathematical expression. If we don’t follow this order then the value which we will get will be incorrect. Here, we can make a mistake by writing the negative power of \[q\] as \[-6\] because it is already negative. But, we have to find the negative of this power, so 6 will be the answer.