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if lnx=1/x, then what is ln1/x=?

 Dec 9, 2015

Best Answer 

 #1
avatar+23278 
+15

if lnx=1/x, then what is ln1/x=?

 

\(\begin{array}{rcll} \ln{(x)} =\frac{1}{x} \qquad \ln{( \frac{1}{x} )} = \ ? \end{array}\)

 

\( \begin{array}{rcll} \ln{( \frac{1}{x} )} &=& \ln{( 1 )} -\ln{( x )} \qquad & | \qquad \ln{( 1 )} = 0\\ &=& 0 -\ln{( x )} \\ &=& -\ln{( x )} \qquad & | \qquad \ln{(x)} =\frac{1}{x}\\ &=& -\frac{1}{x} \\ \mathbf{ \ln{( \frac{1}{x} )} } & \mathbf{=} & \mathbf{ -\frac{1}{x} } \end{array}\)

 

 

excursus

\(\begin{array}{rcll} \ln{(x)} = \frac{1}{x} \qquad x = \ ? \end{array}\)

 

\(\begin{array}{rcll} \ln{(x)} &=& \frac{1}{x} \\ \ln{(x^x)} &=& 1 \qquad & | \qquad e^{()} \\ \mathbf{x^x} &\mathbf{=}& \mathbf{e} \\\\ e^{\ln{(x^x)}}&=& e \qquad \ln{(x^x)} = 1\\ e^{ x\cdot \ln{(x)} } &=& e\\ x\cdot \ln{(x)} &=& 1 \qquad z = \ln{(x)} \\ x\cdot z &=& 1 \qquad e^z = x \\ e^z\cdot z &=& 1 \\\\ z &=& W(1) \\ e^{\ln{(x)}} &=& e^{W(1)}\\ x &=& e^{W(1)} \qquad \text{or}\\\\ x\cdot z &=& 1 \\ x &=& \frac{1}{z} \qquad z = W(1) \\ x &=& \frac{1}{W(1)}\\\\ \end{array}\\ \small{ W(1) = 0.5671432904097838729999686622103555497538157871865125081351310792230457930866\dots\\ x = \frac{1}{W(1)} = 1.7632228343518967102252017769517070804360179866674736\dots\\\\ W(1) = 0.5671432904\dots \text{ is called the omega constant }\\ \text{where } W(x) \text{ is the Lambert W-function } }\)

 

laugh

 Dec 9, 2015
 #1
avatar+23278 
+15
Best Answer

if lnx=1/x, then what is ln1/x=?

 

\(\begin{array}{rcll} \ln{(x)} =\frac{1}{x} \qquad \ln{( \frac{1}{x} )} = \ ? \end{array}\)

 

\( \begin{array}{rcll} \ln{( \frac{1}{x} )} &=& \ln{( 1 )} -\ln{( x )} \qquad & | \qquad \ln{( 1 )} = 0\\ &=& 0 -\ln{( x )} \\ &=& -\ln{( x )} \qquad & | \qquad \ln{(x)} =\frac{1}{x}\\ &=& -\frac{1}{x} \\ \mathbf{ \ln{( \frac{1}{x} )} } & \mathbf{=} & \mathbf{ -\frac{1}{x} } \end{array}\)

 

 

excursus

\(\begin{array}{rcll} \ln{(x)} = \frac{1}{x} \qquad x = \ ? \end{array}\)

 

\(\begin{array}{rcll} \ln{(x)} &=& \frac{1}{x} \\ \ln{(x^x)} &=& 1 \qquad & | \qquad e^{()} \\ \mathbf{x^x} &\mathbf{=}& \mathbf{e} \\\\ e^{\ln{(x^x)}}&=& e \qquad \ln{(x^x)} = 1\\ e^{ x\cdot \ln{(x)} } &=& e\\ x\cdot \ln{(x)} &=& 1 \qquad z = \ln{(x)} \\ x\cdot z &=& 1 \qquad e^z = x \\ e^z\cdot z &=& 1 \\\\ z &=& W(1) \\ e^{\ln{(x)}} &=& e^{W(1)}\\ x &=& e^{W(1)} \qquad \text{or}\\\\ x\cdot z &=& 1 \\ x &=& \frac{1}{z} \qquad z = W(1) \\ x &=& \frac{1}{W(1)}\\\\ \end{array}\\ \small{ W(1) = 0.5671432904097838729999686622103555497538157871865125081351310792230457930866\dots\\ x = \frac{1}{W(1)} = 1.7632228343518967102252017769517070804360179866674736\dots\\\\ W(1) = 0.5671432904\dots \text{ is called the omega constant }\\ \text{where } W(x) \text{ is the Lambert W-function } }\)

 

laugh

heureka Dec 9, 2015
 #2
avatar+105509 
+5

Thanks Heureka,

 

That excursion was really convoluted.

 

Intriguing - but convoluted.    laughindecisionlaugh

 Dec 9, 2015

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