Find the value of \[2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right]\].
Answer
601.5k+ views
Hint:
Here, we will first simplify the given arithmetic expression by using the Negative Exponent rule. Then we will use the BODMAS rule and simplify further to get the required answer. Arithmetic Expression is defined as an expression with numbers and arithmetic operators.
Formula Used:
Negative Exponent Rule: \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Complete step by step solution:
We are given with an Arithmetic Expression \[2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right]\].
Now, by using the Negative Exponent Rule \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\], we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{2} \times \dfrac{1}{4}} \right) \div \dfrac{1}{{{2^2}}}} \right]\]
We will simplify the arithmetic expression by using the BODMAS rule.
Now, by using the BODMAS Rule, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{8}} \right) \div \dfrac{1}{{{2^2}}}} \right]\]
Applying the exponent on the terms, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{8}} \right) \div \dfrac{1}{4}} \right]\]
Now, by dividing the numbers inside the bracket, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\dfrac{4}{8}} \right]\]
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\dfrac{1}{2}} \right]\]
Multiplying the numbers and canceling out the numbers, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 1\]
Therefore, the value of \[2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right]\] is \[1\] .
Note:
We know that the BODMAS rule states that the first operation has to be done which is in the brackets, next the operation applies on the indices or order, then it moves on to the division and multiplication and then using addition and subtraction we will simplify the expression. If addition or subtraction and division or multiplication are in the same calculations, then it has to be done from left to right. An arithmetic expression is defined as an expression with the numbers and the arithmetic operators like plus, minus etc. Negative Exponent rule, which says that the negative exponent in the numerator gets changed to the denominator, then the exponent becomes positive.
Here, we will first simplify the given arithmetic expression by using the Negative Exponent rule. Then we will use the BODMAS rule and simplify further to get the required answer. Arithmetic Expression is defined as an expression with numbers and arithmetic operators.
Formula Used:
Negative Exponent Rule: \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Complete step by step solution:
We are given with an Arithmetic Expression \[2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right]\].
Now, by using the Negative Exponent Rule \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\], we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{2} \times \dfrac{1}{4}} \right) \div \dfrac{1}{{{2^2}}}} \right]\]
We will simplify the arithmetic expression by using the BODMAS rule.
Now, by using the BODMAS Rule, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{8}} \right) \div \dfrac{1}{{{2^2}}}} \right]\]
Applying the exponent on the terms, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\left( {\dfrac{1}{8}} \right) \div \dfrac{1}{4}} \right]\]
Now, by dividing the numbers inside the bracket, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\dfrac{4}{8}} \right]\]
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 2\left[ {\dfrac{1}{2}} \right]\]
Multiplying the numbers and canceling out the numbers, we get
\[ \Rightarrow 2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right] = 1\]
Therefore, the value of \[2\left[ {\left( {{2^{ - 1}} \times {4^{ - 1}}} \right) \div {2^{ - 2}}} \right]\] is \[1\] .
Note:
We know that the BODMAS rule states that the first operation has to be done which is in the brackets, next the operation applies on the indices or order, then it moves on to the division and multiplication and then using addition and subtraction we will simplify the expression. If addition or subtraction and division or multiplication are in the same calculations, then it has to be done from left to right. An arithmetic expression is defined as an expression with the numbers and the arithmetic operators like plus, minus etc. Negative Exponent rule, which says that the negative exponent in the numerator gets changed to the denominator, then the exponent becomes positive.
Recently Updated Pages
Match columnI with columnII and choose the correct class 12 biology NEET_UG

Match columnI with columnII and choose the correct class 12 biology NEET_UG

Match columnI with columnII and choose the correct class 12 biology NEET_UG

Which plant will lose its economic value if its fruits class 12 biology NEET_UG

The given figure shows two endocrine glands marked class 11 biology NEET_UG

Human insulin is being commercially produced from a class 12 biology NEET_UG

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

One cusec is equal to how many liters class 8 maths CBSE

Give me the opposite gender of Duck class 8 english CBSE

Answer the following in a phrase or sentence Why did class 8 english CBSE

Full form of STD, ISD and PCO


