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If ${{\text{x}}^{\text{y}}}{\text{ = }}{{\text{y}}^{\text{x}}}$and x = 2y, then find the values of x and y (x,y >0)
A. x = 4, y = 2
B. x = 3, y = 2
C. x = 1, y = 1
D. none of these.

Answer
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Hint – In order to solve this problem put the value of x in the given equation and solve to find the value of y. Then put the value of y in which x is present then solve for x. Doing this will make your problem solved.

Complete step-by-step answer:
The given equations are :
$ \to {{\text{x}}^{\text{y}}}{\text{ = }}{{\text{y}}^{\text{x}}}$ ……(1)
x = 2y ……(2)
Taking log both sides in equation (1) we get,
$ \to {\text{log}}{{\text{x}}^{\text{y}}}{\text{ = log}}{{\text{y}}^{\text{x}}}$
Solving it further we get,
$
  \because \log {a^b} = a\log b \\
   \to {\text{ylogx = xlogy}} \\
   \to \dfrac{{{\text{logx}}}}{{\text{x}}}{\text{ = }}\dfrac{{{\text{logy}}}}{{\text{y}}} \\
$
On putting the value of x from (1) in the above equation we will get the new equation as
$ \to \dfrac{{{\text{log2y}}}}{{{\text{2y}}}}{\text{ = }}\dfrac{{{\text{logy}}}}{{\text{y}}}$
Simplifying the above equation we get,
$ \to {\text{log2y = 2logy}}$
$ \to {\text{log2y - 2logy = 0}}$
As we know ${\text{logab = loga}}\,{\text{ + logb}}$applying the same in above equation we get,
$
   \to {\text{log2 + logy - 2logy = 0}} \\
   \to {\text{log2 - logy = 0}} \\
   \to {\text{logy = log2}} \\
   \to {\text{y = 2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(}}\because {\text{loga = logb}} \to {\text{a = b}}) \\
$
On putting the value of y in equation (2) we will get the value of x as,
$ \to $x = 2(2)
$ \to $x = 4
Hence the value of y is 2 and that of x is 4.
So, the correct option is (A).

Note – Whenever you face this type of problem then try to use the concepts of logarithms it will make your problem a bit easier to solve. Here we have taken log and solved the equation using properties of log to reach the right answer.