ln(1/(x-1))=1

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Solution for ln(1/(x-1))=1 equation: 


D( x )

1/(x-1) <= 0

x-1 = 0

1/(x-1) <= 0

1/(x-1) <= 0

( 1 )

1 <= 0

x belongs to the empty set

( x-1 )

x-1 <= 0 // + 1

x <= 1

	(-oo:1)	(1:+oo)
1	+	+
x-1	-	+


x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

x in (1:+oo)

ln(1/(x-1)) = 1 // - 1

ln(1/(x-1))-1 = 0

ln(1/(x-1))-1 = 0 // + 1

ln(1/(x-1)) = 1

ln(1/(x-1)) = ln(e^1)

1/(x-1) = e^1

1/(x-1)-e^1 = 0

1/(x-1)-e = 0

1/(x-1)+(-e*(x-1))/(x-1) = 0

1-e*(x-1) = 0

e-e*x+1 = 0

(e-e*x+1)/(x-1) = 0

(e-e*x+1)/(x-1) = 0 // * x-1

e-e*x+1 = 0

e-e*x+1 = 0 // - e+1

-e*x = -(e+1) // : -e

x = (-(e+1))/(-e)

x = (e+1)/e

x = (e+1)/e

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