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# Simplify the expression which is given by ${\left( {\dfrac{{2{x^3}{y^2}}}{{3{x^2}{y^5}}}} \right)^4} \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right)$. Verified
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Hint- Here, we will proceed by using the concepts that when two numbers having the same bases are multiplied and divided, their powers will be added and subtracted respectively in order to simplify the given expression.

Let us denote the given expression by a i.e., $a = {\left( {\dfrac{{2{x^3}{y^2}}}{{3{x^2}{y^5}}}} \right)^4} \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right)$
$\Rightarrow a = \dfrac{{{{\left( {2{x^3}{y^2}} \right)}^4}}}{{{{\left( {3{x^2}{y^5}} \right)}^4}}} \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right) \\ \Rightarrow a = \dfrac{{{{\left( 2 \right)}^4}{{\left( {{x^3}} \right)}^4}{{\left( {{y^2}} \right)}^4}}}{{{{\left( 3 \right)}^4}{{\left( {{x^2}} \right)}^4}{{\left( {{y^5}} \right)}^4}}} \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right){\text{ }} \to {\text{(1)}} \\$
As we know that for any number a, ${\left( {{a^b}} \right)^c} = {a^{b \times c}}$
Also, ${\left( 2 \right)^4} = 16$ and ${\left( 3 \right)^4} = 81$
By using the formula ${\left( {{a^b}} \right)^c} = {a^{b \times c}}$ and putting ${\left( 2 \right)^4} = 16$, ${\left( 3 \right)^4} = 81$, equation (1) becomes
$\Rightarrow a = \left( {\dfrac{{16{x^{3 \times 4}}{y^{2 \times 4}}}}{{81{x^{2 \times 4}}{y^{5 \times 4}}}}} \right) \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right) \\ \Rightarrow a = \left( {\dfrac{{16{x^{12}}{y^8}}}{{81{x^8}{y^{20}}}}} \right) \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right) \\ \Rightarrow a = \left[ {\dfrac{{\left( {16{x^{12}}{y^8}} \right)\left( {{x^2}{y^2}} \right)}}{{\left( {81{x^8}{y^{20}}} \right)\left( {12} \right)}}} \right] \\ \Rightarrow a = \left[ {\dfrac{{4\left( {{x^{12}} \times {x^2}} \right)\left( {{y^8} \times {y^2}} \right)}}{{\left( {81{x^8}{y^{20}}} \right)\left( 3 \right)}}} \right]{\text{ }} \to {\text{(2)}} \\$
Also we know that when two numbers such that their bases are same are multiplied with each other then, their powers will be added i.e., ${a^b} \times {a^c} = {a^{b + c}}$.
Using the above concept, equation (2) becomes
$\Rightarrow a = \left[ {\dfrac{{4\left( {{x^{12 + 2}}} \right)\left( {{y^{8 + 2}}} \right)}}{{\left( {81{x^8}{y^{20}}} \right)\left( 3 \right)}}} \right] \\ \Rightarrow a = \left[ {\dfrac{{4\left( {{x^{14}}} \right)\left( {{y^{10}}} \right)}}{{243{x^8}{y^{20}}}}} \right] \\ \Rightarrow a = \left[ {\left( {\dfrac{4}{{243}}} \right)\left( {\dfrac{{{x^{14}}}}{{{x^8}}}} \right)\left( {\dfrac{{{y^{10}}}}{{{y^{20}}}}} \right)} \right]{\text{ }} \to {\text{(3)}} \\$
Also we know that when two numbers such that their bases are same are divided then, their powers will be subtracted i.e., $\left( {\dfrac{{{a^b}}}{{{a^c}}}} \right) = {a^{b - c}}$.
Using the above concept, equation (3) becomes
$\Rightarrow a = \left[ {\left( {\dfrac{4}{{243}}} \right)\left( {{x^{14 - 8}}} \right)\left( {{y^{10 - 20}}} \right)} \right] \\ \Rightarrow a = \left( {\dfrac{4}{{243}}} \right)\left( {{x^6}} \right)\left( {{y^{ - 10}}} \right) \\ \Rightarrow a = \dfrac{{4{x^6}}}{{243{y^{10}}}} \\$
Therefore, the given expression ${\left( {\dfrac{{2{x^3}{y^2}}}{{3{x^2}{y^5}}}} \right)^4} \times \left( {\dfrac{{{x^2}{y^2}}}{{12}}} \right)$ is simplified to $\dfrac{{4{x^6}}}{{243{y^{10}}}}$.

Note- In this particular problem, $\left( {{x^{12}} \times {x^2}} \right) = {x^{14}}$ because here both the terms ${x^{12}}$ and ${x^2}$ have the same base (i.e., x) and are multiplied with each other so their powers will be added. Also, $\left( {{y^8} \times {y^2}} \right) = {y^{10}}$ because here both the terms ${y^8}$ and ${y^2}$ have the same base (i.e., y) and are multiplied with each other so their powers will be added.

Last updated date: 30th Sep 2023
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