Question

X^y=y^x and x=2y prove that y=2.

Answer 1

Answer:

${x}^{y} = {y}^{x} \: - - - (1)$

$x = 2y \: - - - (2)$

Substitute (2) into (1):

${2y}^{y} = {y}^{2y}$

Log both sides:

$lg(2 {y}^{y} ) = lg(y^{2y} )$

$y \: lg(2y) = 2y(lg \: y)$

$y \: lg(2) + y \: lg(y) = 2y \: lg(y)$

$y \: lg(2) = 2y \: lg(y) - y \: lg(y)$

$y \: lg(2) = y \: lg(y)$

Divide by y on both sides:

$lg(2) = lg(y)$

${10}^{lg(2)} = {10}^{lg(y)}$

2= y

y= 2 (proved)

Explanation:

• step 5: lg(m)^n= nlg(m)

• Step 6: lg(mn)= lg(m) +lg(n)