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Question: Find the values of x and y if, \({{y}^{x}}={{x}^{y}}\) and x = 2y....

Find the values of x and y if, yx=xy{{y}^{x}}={{x}^{y}} and x = 2y.

Explanation

Solution

Hint: To solve problems related to algebra different techniques may be employed. Particularly for this problem we are provided with two equations.For first equation use logarithm property and then substitute second equation in first , we get the final expression in one variable and simplify it to get the values of x and y.

Complete step-by-step answer:
We proceed by applying log to both sides and then by using the second equation we get the final expression in one variable which yields a particular value for that variable. And if the first variable is solved then we can do the same to the second involved variable.

First, we label the equation with a particular number such that yx=xy…(1){{y}^{x}}={{x}^{y}}\ldots (1).
Similarly, label x=2y…(2)x=2y\ldots (2).
The basic log property which is useful in solving this problem:
log⁡ab=blog⁡a\log {{a}^{b}}=b\log a
Considering the equation (1) and taking log of both sides we get,
log⁡yx=log⁡xy xlog⁡y=ylog⁡x log⁡yy=log⁡xx…(3) \begin{aligned} & \log {{y}^{x}}=\log {{x}^{y}} \\\ & x\log y=y\log x \\\ & \dfrac{\log y}{y}=\dfrac{\log x}{x}\ldots (3) \\\ \end{aligned}
Equation (3) would be useful in evaluation of the result.
Considering equation (2) we get, x=2yx=2y.
Now, putting the value of equation (2) in equation (3) we get,
log⁡yy=log⁡2y2y\dfrac{\log y}{y}=\dfrac{\log 2y}{2y}
Now, by using another property of log i.e. log⁡(a⋅b)=log⁡a+log⁡b\log (a\cdot b)=\log a+\log b.
So, log2y can be expanded using the above identity.
Hence, the evaluated result yields:
log⁡yy=log⁡2+log⁡y2y log⁡2+log⁡ylog⁡y=2yy log⁡2log⁡y+1=2 log⁡2log⁡y=1 ∴log⁡y=log⁡2…(4) \begin{aligned} & \dfrac{\log y}{y}=\dfrac{\log 2+\log y}{2y} \\\ & \dfrac{\log 2+\log y}{\log y}=\dfrac{2y}{y} \\\ & \dfrac{\log 2}{\log y}+1=2 \\\ & \dfrac{\log 2}{\log y}=1 \\\ & \therefore \log y=\log 2\ldots (4) \\\ \end{aligned}
For taking antilog of any log quantity we get, elog⁡a=a{{e}^{\log a}}=a
Now, taking the antilog of equation (4) we get y as 2.
Now, placing the obtained value of y in equation (2) we get x as 4.
Hence, x = 4 and y = 2.

Note: This problem can also be solved by a hit and trial method using various combinations of x and y. Like, when y = 1 we get x = 2 which would be incorrect. But when we consider y = 2 we obtain x = 4. So, this satisfies our power equation and hence it is correct.
The key step is a stepwise method of solving log by using appropriate property.