-2=-ln(-ln(1-1/x))

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Solution for -2=-ln(-ln(1-1/x)) equation: 


D( x )

x = 0

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

1-(1/x) <= 0

x = 0

x = 0

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

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

-ln(1-x^-1) <= 0

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

-ln(1-x^-1) <= 0

-ln(1-x^-1) in (-oo:0>

-ln(1-x^-1) <= 0 // * -1

-ln(1-x^-1) >= ln(e^0)

1-x^-1 >= e^0

1-x^-1-e^0 >= 0

-1*x^-1 >= 0

-1*x^-1 >= 0 // * -1

1*x^-1 <= 0

1*x^-1 <= 0 // : 1

x^-1 <= 0/1

x^-1 <= 0

1/(x^1) <= 0

x <> 0

1/(x^1) <= 0 // * x^2

(x^2)/(x^1) <= 0

x^1 <= 0

x <= 0

x in (-oo:0)

1-(1/x) <= 0

1-(1/x) <= 0

1-x^-1 <= 0

-1*x^-1 <= -1

-1*x^-1 <= -1 // * -1

1*x^-1 >= 1

1*x^-1 >= 1 // : 1

x^-1 >= 1/1

x^-1 >= 1

1/(x^1) >= 1

x <> 0

1/(x^1) >= 1 // * x^2

(x^2)/(x^1) >= 1*x^2

x^1 >= 1*x^2

x-x^2 >= 0

x^1*(1-x) >= 0

x < 0

-(abs(x))^1*(1*(abs(x))^1+1) >= 0 // : -(abs(x))^1

1*(abs(x))^1+1 <= 0

1*(abs(x))^1 <= -1 // : 1

(abs(x))^1 <= -1/1

(abs(x))^1 <= -1 // ^ 1/1

abs(x) <= (-1)^(1/1)

abs(x) <= -1

x belongs to the empty set

x < 0

x > 0

x^1*(1-(1*x^1)) >= 0 // : x^1

1-(1*x^1) >= 0

1 >= 1*x^1 // : 1

1/1 >= x^1 // ^ 1/1

(1/1)^(1/1) >= x

x in (0:1>

x > 0

x in (0:1>

x in (1:+oo)

-2 = -ln(-ln(1-(1/x))) // + ln(-ln(1-(1/x)))

ln(-ln(1-(1/x)))-2 = 0

ln(-ln(1-x^-1))-2 = 0

ln(-ln(1-x^-1))-2 = 0 // + 2

ln(-ln(1-x^-1)) = 2

ln(-ln(1-x^-1)) = ln(e^2)

-ln(1-x^-1) = e^2

-ln(1-x^-1)-e^2 = 0

-ln(1-x^-1)-e^2 = 0 // + e^2

-ln(1-x^-1) = e^2

-ln(1-x^-1) = e^2 // * -1

ln(1-x^-1) = ln(e^(-e^2))

1-x^-1 = e^(-e^2)

-1*x^-1 = -(1-e^(-e^2)) // : -1

x^-1 = 1-e^(-e^2)

-1 < 0

1/(x^1) = 1-e^(-e^2) // * x^1

1 = (1-e^(-e^2))*x^1 // : 1-e^(-e^2)

1/(1-e^(-e^2)) = x^1

x = 1/(1-e^(-e^2))

x = 1/(1-e^(-e^2))

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