Consider equations ` x^(log_(y)x) = 2 and y^(log_(x)y) = 16`.
The value of x is

Options

A. `2^(root(3)2)`
B. ` 2^(root(3)4)`
C. ` 2^(root(3)64)`
D. ` 2 root(3)256)`

Answer

B

Explanation

Let ` log_(y) x = 1`
Then ` x = y^(t)` …(1)
Now, ` x^(log_(y) x) =2` becomes
` x^(t) = 2`
`rArr x = 2^(1//t)` …(2)
And `y^(log_(x)y) = 16` becomes
` y^(1//t) = 2^(4)`
` rArr y = 2^(4//t)` ….(3)
Putting the values of x and y in (1), we get
` 2^(1//t) = 2^(4t^(2))`
` rArr 4t^(3) = 1`
` :. t = (1/4)^(1//3)` ....(4)
Using (4) and (2), we get ` x = (2)^((4)^(1//3)) = 2^(root(3)4)`
Using (4) and (3), we get ` y = (2)^((4)^(2//3)) = 2^(root(3)16)`