# 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.

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.

Last updated date: 18th Sep 2023

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