Question

\[\text{If }{{x}^{y}}={{y}^{x}},\text{ then }{{\left( \dfrac{x}{y} \right)}^{\dfrac{x}{y}}}\text{ equals}\]
(a) \[{{x}^{\dfrac{x}{y}}}\]
(b) \[{{x}^{\left( \dfrac{x}{y} \right)-1}}\]
(c) \[{{x}^{\dfrac{y}{x}}}\]
(d) \[{{x}^{\left( \dfrac{y}{x} \right)-1}}\]

Answer 1

Hint: First of all, consider the given equation and take log on both its sides and from that find the value of y. Now, substitute this value of y in the expression asked in the question to get the required answer.

Complete step-by-step answer:
We are given that \[{{x}^{y}}={{y}^{x}}\]. We have to find the value of \[{{\left( \dfrac{x}{y} \right)}^{\dfrac{x}{y}}}\]. Let us consider the equation given in the question.
\[{{x}^{y}}={{y}^{x}}\]
By taking log on both the sides of the above equation, we get,
\[\log {{x}^{y}}=\log {{y}^{x}}\]
We know that \[\log {{a}^{b}}=b\log a\]. By using this, we get,
\[y\log x=x\log y\]
By dividing x log x on both the sides of the equation, we get,
\[\dfrac{y\log x}{x\log x}=\dfrac{x\log y}{x\log x}\]
By canceling the like terms of the above equation, we get,
\[\dfrac{y}{x}=\dfrac{\log y}{\log x}\]
We know that \[\dfrac{\log a}{\log b}={{\log }_{b}}a\], by using this, we get,
\[{{\log }_{x}}y=\dfrac{y}{x}\]
We know that if \[{{\log }_{b}}a=t,\text{ then }a={{\left( b \right)}^{t}}\], by using this, we get,
\[y={{\left( x \right)}^{\dfrac{y}{x}}}....\left( i \right)\]
Now, let us consider the expression asked in the question.
\[E={{\left( \dfrac{x}{y} \right)}^{\dfrac{x}{y}}}\]
By substituting the value of y in the above expression from equation (i), we get,
\[E={{\left( \dfrac{x}{{{\left( x \right)}^{\dfrac{y}{x}}}} \right)}^{\dfrac{x}{y}}}\]
We know that \[{{\left( \dfrac{a}{b} \right)}^{x}}=\dfrac{{{a}^{x}}}{{{b}^{x}}}\]. By using this, we get,
\[E=\dfrac{{{x}^{\dfrac{x}{y}}}}{{{x}^{\dfrac{y}{x}.\dfrac{x}{y}}}}\]
\[E=\dfrac{{{x}^{\dfrac{x}{y}}}}{{{x}^{1}}}\]
We know that \[\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}\]. By using this, we get,
\[E={{\left( x \right)}^{\dfrac{x}{y}-1}}\]
Hence, we get the value of \[{{\left( \dfrac{x}{y} \right)}^{\dfrac{x}{y}}}={{\left( x \right)}^{\dfrac{x}{y}-1}}\]
Hence, option (b) is correct.

Note: In the above question, students can find the value of y without using the log as well. That is they can raise both sides of the given equation by \[\dfrac{1}{x}\] as follows:
\[{{x}^{y}}={{y}^{x}}\]
\[{{\left( {{x}^{y}} \right)}^{\dfrac{1}{x}}}={{\left( {{y}^{x}} \right)}^{\dfrac{1}{x}}}\]
\[{{x}^{\dfrac{y}{x}}}={{y}^{\dfrac{x}{x}}}\]
\[y={{x}^{\dfrac{y}{x}}}\]
Hence, students are advised to learn both the methods to clearly and easily solve these types of questions. Also, students must see the options first and then accordingly solve the question. Like in the above question, options are given by taking base x. So, we can have also do it in that way