Question

Simplify the given exponent : $\dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}}$
A.$\dfrac{3}{2}$
B.$\dfrac{5}{2}$
C.$ - \dfrac{1}{2}$
D.$\dfrac{1}{2}$

Answer 1

Hint: Here, we will first simplify the exponent of the given terms by using the property of exponents. Then we will use basic mathematical operations like subtraction and multiplication to further simplify the expression further and get the required value of the expression.

Formula Used:
${a^{m + n}} = {a^m} \times {a^n}$

Complete step-by-step answer:
To solve the expression $\dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}}$ we will use the property of exponents.
This is because both the numerator as well as the denominator is having the product of exponential numbers. Hence, using the laws of exponents will help us to solve this question and find the required answer to the given fraction.
Thus, using the identity ${a^{m + n}} = {a^m} \times {a^n}$, we can write the given expression as:
$\dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}} = \dfrac{{16 \times {2^n} \times 2 - 4 \times {2^n}}}{{16 \times {2^n} \times {2^2} - 2 \times {2^n} \times {2^2}}}$
Taking ${2^n}$ common from both the numerator as well as the denominator, we get,
$ \Rightarrow \dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}} = \dfrac{{{2^n}\left( {16 \times 2 - 4} \right)}}{{{2^n}\left( {16 \times {2^2} - 2 \times {2^2}} \right)}}$
Multiplying the terms in the numerator and applying the exponents on terms in the denominator, we get
$ \Rightarrow \dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}} = \dfrac{{32 - 4}}{{16 \times 4 - 2 \times 4}}$
Again multiplying the terms in the denominator, we get
$ \Rightarrow \dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}} = \dfrac{{32 - 4}}{{64 - 8}}$
Subtracting the terms, we get
$ \Rightarrow \dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}} = \dfrac{{28}}{{56}} = \dfrac{1}{2}$
Therefore, the value of $\dfrac{{16 \times {2^{n + 1}} - 4 \times {2^n}}}{{16 \times {2^{n + 2}} - 2 \times {2^{n + 2}}}}$ is $\dfrac{1}{2}$
Hence, option D is the correct answer.

Note: An expression that represents the repeated multiplication of the same number is known as power. When a number is written with power then the power becomes the exponent of that particular number. It shows how many times that particular number will be multiplied by itself. Hence, whenever we are given the multiplication of the same numbers, then we can express that number with an exponent and vice-versa.
Here, we have used the BODMAS rule, according to which we do subtraction after doing multiplication. Hence, first of all we have multiplied the numbers which had a multiplication sign between them and then, we have done the subtraction as per the requirement of the question.