Question

Chapter - Laws of Indices

Solve :

$x^{y}=y^{x},\:x^{2}=y^{3}\:(x,y>0).$​

Answer 1

$x^{y}=y^{x} \\ {x}^{2y} = {y}^{2x} \: \: \: [ \: by \: squaring \: both \: side\: ] \\ {y}^{3y} = {y}^{2x} \\ 3y = 2x \\ \frac{x}{y} = \frac{2}{3}$

Answer 2

Solution:

Given,

x^y = y^x

x^2 = y^3

where ( x, y > 0 )

Now, solve

x^y = y^x

=> x^2y = y^2x [Reason: Squaring both sides]

=> y^3y = y^2x

Now, take the exponents of both

3y = 2x

=> x/y = 2/3

Therefore, x = 2 and y = 3