Question

Find the positive solutions of the system of equations \[{x^{x + y}} = {y^n}\] and \[{y^{x + y}} = {x^{2n}}{y^n}\] where \[n > 0\].

Answer 1

Hint: Here we will proceed by multiplying the given equations and we will find the relations between the variables using the algebraic formula. Further we will use a substitution method to get the required positive values of the given system of equations.

Complete step-by-step answer:
Given system of equations are \[{x^{x + y}} = {y^n}...................................................................\left( 1 \right)\]
And \[{y^{x + y}} = {x^{2n}}{y^n}....................................................................\left( 2 \right)\]
Multiplying the equations (1) and (2), we have
\[ \Rightarrow {x^{x + y}} \times {y^{x + y}} = {y^n} \times {x^{2n}}{y^n}\]
Since, the powers are equal we can multiply the bases and we get
\[
   \Rightarrow {\left( {xy} \right)^{x + y}} = {y^{2n}} \times {x^{2n}} \\
   \Rightarrow {\left( {xy} \right)^{x + y}} = {\left( {xy} \right)^{2n}} \\
\]
Since, the bases are equal we can equate the powers. By equating the powers, we get
\[ \Rightarrow x + y = 2n.......................................................\left( 3 \right)\]
Substituting equation (3) in equation (1), we have
\[ \Rightarrow {x^{2n}} = {y^n}\]
By using the formula, \[{a^{mn}} = {\left( {{a^m}} \right)^n}\] we get
\[ \Rightarrow {\left( {{x^2}} \right)^n} = {y^n}\]
Since, the bases are equal we can equate the powers. By equating the powers, we get
\[\therefore {x^2} = y...................................................\left( 4 \right)\]
Substituting equation (4) in equation (3), we have
\[
   \Rightarrow x + {x^2} = 2n \\
   \Rightarrow {x^2} + x - 2n = 0 \\
\]
We know that the roots of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
So, the roots of the equation \[{x^2} + x - 2n = 0\] are given by
\[
   \Rightarrow x = \dfrac{{ - \left( 1 \right) \pm \sqrt {{1^2} - 4\left( 1 \right)\left( { - 2n} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{ - 1 \pm \sqrt {1 + 8n} }}{2} \\
\]
Since, \[n > 0\] the positive values of \[x = \dfrac{{ - 1 + \sqrt {1 + 8n} }}{2},\dfrac{{ - 1 - \sqrt {1 + 8n} }}{2}\].
Substituting \[x = \dfrac{{ - 1 \pm \sqrt {1 + 8n} }}{2}\] in equation (3), we get
\[
   \Rightarrow \dfrac{{ - 1 \pm \sqrt {1 + 8n} }}{2} + y = 2n \\
   \Rightarrow y = 2n - \dfrac{{ - 1 \pm \sqrt {1 + 8n} }}{2} \\
   \Rightarrow y = \dfrac{{4n - 1 \pm \sqrt {1 + 8n} }}{2} \\
\]
Since, \[n > 0\] the positive values of \[y = \dfrac{{4n - 1 + \sqrt {1 + 8n} }}{2},\dfrac{{4n - 1 - \sqrt {1 + 8n} }}{2}\].
Thus, the required positive values are \[x = \dfrac{{ - 1 + \sqrt {1 + 8n} }}{2},\dfrac{{ - 1 - \sqrt {1 + 8n} }}{2}\] and \[y = \dfrac{{4n - 1 + \sqrt {1 + 8n} }}{2},\dfrac{{4n - 1 - \sqrt {1 + 8n} }}{2}\] for all values of \[n > 0\].

Note: The roots of a quadratic equation of the form \[a{x^2} + bx + c = 0\] is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Here we have used a substitution method to find the unknown values of \[x,y\] in terms of \[n\]. Always remember that whenever the powers are equal then we can equate the bases and vice-versa.