Question

Simplify: \[\dfrac{{{7^{2n + 3}}{{\left( {49} \right)}^{n + 2}}}}{{{{\left( {{{\left( {343} \right)}^{n + 1}}} \right)}^{{{\left({{2/3}}\right)}
}}}}}\]

Answer 1

Hint:
Here we will use basic rules of exponentials to simplify the problem. For that, we will first convert 49 into exponentials and 343 into exponentials. Then we will use power rules for exponents to simplify the expression. We will use the product rule of exponentials to multiply the exponents. After applying all the rules, we will get the required answer.

Complete step by step solution:
Let \[S\] be the simplified value of \[\dfrac{{{7^{2n + 3}}{{\left( {49} \right)}^{n + 2}}}}{{{{\left( {{{\left( {343} \right)}^{n + 1}}} \right)}^{\dfrac{2}{3}}}}}\] .
Therefore,
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}}{{\left( {49} \right)}^{n + 2}}}}{{{{\left( {{{\left( {343} \right)}^{n + 1}}} \right)}^{\dfrac{2}{3}}}}}\] …….. \[\left( 1 \right)\]
First we will convert 49 into exponential. We know, \[49 = {7^2}\], thus we will substitute the value of 49 in equation \[\left( 1 \right)\]. Therefore, we get
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}}{{\left( {{7^2}} \right)}^{n + 2}}}}{{{{\left( {{{\left( {343} \right)}^{n + 1}}} \right)}^{\dfrac{2}{3}}}}}\] ……… \[\left( 2 \right)\]
Now we will convert 343 into exponential. We know, \[343 = {7^3}\], thus we will substitute the value of 343 in equation \[\left( 2 \right)\].
So,
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}}{{\left( {{7^2}} \right)}^{n + 2}}}}{{{{\left( {{{\left( {{7^3}} \right)}^{n + 1}}} \right)}^{\dfrac{2}{3}}}}}\]
Using power rules for exponentials, we get
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}} \times {7^{2 \times \left( {n + 2} \right)}}}}{{{{\left( {{7^3}^{ \times \left( {n + 1} \right)}} \right)}^{\dfrac{2}{3}}}}}\]
Again using the power rule for exponential in denominator, we get
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}} \times {7^{2 \times \left( {n + 2} \right)}}}}{{{7^3}{{^{ \times \left( {n + 1} \right) \times }}^{\dfrac{2}{3}}}}}\]
On further simplification, we get
\[ \Rightarrow S = \dfrac{{{7^{2n + 3}} \times {7^{2n + 4}}}}{{{7^{2n + 2}}}}\]
We will multiply the exponents in the numerator. So, we get
\[ \Rightarrow S = \dfrac{{{7^{2n + 3 + 2n + 4}}}}{{{7^{2n + 2}}}}\]
Simplifying the terms in power, we get
\[ \Rightarrow S = \dfrac{{{7^{4n + 7}}}}{{{7^{2n + 2}}}}\]
Dividing the exponentials by subtracting their powers, we get
\[ \Rightarrow S = {7^{4n + 7 - 2n - 2}}\]
On further simplification, we get
\[ \Rightarrow S = {7^{2n + 5}}\]

Hence, the simplified value of \[\dfrac{{{7^{2n + 3}}{{\left( {49} \right)}^{n + 2}}}}{{{{\left( {{{\left( {343} \right)}^{n + 1}}} \right)}^{\dfrac{2}{3}}}}}\] is \[{7^{2n + 5}}\].

Note:
We have used the basic rule of exponentials here. Power of power rule for exponential means when we raise exponential to another power then the resultant power of the exponential will be the product of those two exponents. Similarly, when we take product of two exponentials with the same base then we simply add the exponents and when we take the quotient of two exponentials with the same base then we simply subtract the exponents.