ln(x/0,11)=-0,0283

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Solution for ln(x/0,11)=-0,0283 equation: 


D( x )

x/0.11 <= 0

x/0.11 <= 0

x/0.11 <= 0

9.09090909*x+0 <= 0 // - 0

9.09090909*x <= 0 // : 9.09090909

x <= 0/9.09090909

x <= 0

x in (0:+oo)

ln(x/0.11) = -0.0283 // + 0.0283

ln(x/0.11)+0.0283 = 0

ln(x/0.11)+0.0283 = 0 // - 0.0283

ln(x/0.11) = -0.0283

ln(x/0.11) = ln(e^-0.0283)

x/0.11 = e^-0.0283

x/0.11-e^-0.0283 = 0

9.09090909*x-e^-0.0283 = 0 // + e^-0.0283

9.09090909*x = e^-0.0283 // : 9.09090909

x = (e^-0.0283)/9.09090909

x = 0.11*e^-0.0283

x = 0.11*e^-0.0283

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