Math miscellaneousHard

Question

If x = y (log y - log x + 1), then the solution of the equation is

Options

A.log = cx
B.x log = cy
C.log = cx
D.log = cy

Solution

= y(log y - log x + 1)

Put y = vx

⇒   v + = (log v + 1)
= v log v
⇒  
put log v = z
dv = dz
⇒  
ln z = ln x + ln c
z = cx
log v = cx
log = cx

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