10^x=x^10.Then find the value of x

10^x=x^10.Then find the value of x

1. \(x > 0 \)

\(\begin{array}{rcll} 10^x &=& x^{10} \qquad & | \qquad \ln() \\ \ln( 10^x )&=& \ln( x^10 ) \\ \boxed{~ \begin{array}{rcll} x \ln( 10 )&=& 10 \ln( x ) \end{array} ~}\\ x \frac{ \ln( 10 )} {10} &=& \ln( x ) \qquad & | \qquad e^{()} \\ e^{x \frac{ \ln( 10 )} {10} } &=& e^{ \ln( x ) } \\ e^{x \frac{ \ln( 10 )} {10} } &=& x \\ \frac{ 1 } { e^{x \frac{ \ln( 10 )} {10} } } &=& \frac{1}{x} \\ e^{ -x \frac{ \ln( 10 )} {10} } &=& \frac{1}{x} \qquad & | \qquad \cdot (-1) \\ - e^{ -x \frac{ \ln( 10 )} {10} } &=& -\frac{1}{x} \qquad & | \qquad \cdot x \\ -x\cdot e^{ -x \frac{ \ln( 10 )} {10} } &=& -1 \qquad & | \qquad \cdot \frac{ \ln( 10 )} {10}\\ \boxed{~ \begin{array}{rcll} -x\cdot\frac{ \ln( 10 )} {10}\cdot e^{ -x \frac{ \ln( 10 )} {10} } &=& -\frac{ \ln( 10 )} {10} \\ \end{array} ~}\\ z &=& -x\cdot\frac{ \ln( 10 )} {10} \\ z\cdot e^z &=& -\frac{ \ln( 10 )} {10} \\ z &=& W(-\frac{ \ln( 10 )} {10} ) \\ -x\cdot \frac{ \ln( 10 )} {10} &=& W(-\frac{ \ln( 10 )} {10})\\ x &=& - \frac{10}{ \ln( 10 )} \cdot W(-\frac{ \ln( 10 )} {10})\\ \end{array}\)

\(\begin{array}{rcll} x &=& -\frac{10}{\ln(10)}\cdot W(-\frac{\ln(10)}{10}) \\ x &=& -4.34294481903\dots \cdot W(-0.23025850930\dots) \\ x &=& -4.34294481903\dots \cdot ( -0.31575086292349132953423736459398378904700244670451503991\dots)\\ x &=& 1.371288574238623536861362106299689958842854404842257070408\dots \end{array}\)

2. \(x<0\)

\(x = -\bar{x}\\ \begin{array}{rcll} 10^{ -\bar{x} } &=& (-\bar{x})^{10} \\ 10^{ -\bar{x} } &=& \bar{x}^{10} \qquad & | \qquad \ln() \\ \ln( 10^{ -\bar{x} } ) &=& \ln(\bar{x})^{10}\\ \boxed{~ \begin{array}{rcll} -\bar{x} \cdot \ln( 10 ) &=& 10 \ln(\bar{x}) \end{array} ~}\\ -\bar{x} \frac{ \ln( 10 )} {10} &=& \ln(\bar{x}) ) \qquad & | \qquad e^{()} \\ e^{ -\bar{x} \frac{ \ln( 10 )} {10} } &=& e^{ \ln(\bar{x}) ) } \\ e^{ -\bar{x} \frac{ \ln( 10 )} {10} } &=& \bar{x} \\ \frac{1} { e^{ \bar{x} \frac{ \ln( 10 )} {10} } } &=& \bar{x} \\ \bar{x} \cdot e^{ \bar{x} \frac{ \ln( 10 )} {10} } &=& 1 \qquad & | \qquad \cdot \frac{ \ln( 10 )} {10}\\ \boxed{~ \begin{array}{rcll} \bar{x} \cdot \frac{ \ln( 10 )} {10} \cdot e^{ \bar{x} \frac{ \ln( 10 )} {10} } &=& \frac{ \ln( 10 )} {10} \end{array} ~}\\ z &=& \bar{x}\cdot\frac{ \ln( 10 )} {10} \\ z\cdot e^z &=& \frac{ \ln( 10 )} {10} \\ z &=& W(\frac{ \ln( 10 )} {10}) \\ \bar{x}\cdot\frac{ \ln( 10 )} {10} &=& W(\frac{ \ln( 10 )} {10}) \\\\ \bar{x} &=& \frac{10}{ \ln( 10 )} \cdot W(\frac{ \ln( 10 )} {10}) \\ x &=&-\bar{x}\\ x &=& -\frac{10}{ \ln( 10 )} \cdot W(\frac{ \ln( 10 )} {10}) \\ \end{array}\)

\(\begin{array}{rcll} x &=& -\frac{10}{\ln(10)}\cdot W(\frac{\ln(10)}{10}) \\ x &=& -4.34294481903\dots \cdot W(0.23025850930\dots) \\ x &=& -4.34294481903\dots \cdot (0.190348104808510145602603466195537000486106560139470579285\dots)\\ x &=& -0.826671315590777916259325344655692315108945325325260676448\dots \end{array}\)

Lambert W Function Calculator see: http://www.had2know.com/academics/lambert-w-function-calculator.html

There are two positive solutions that can be obtained numerically as follows:

There is also a negative solution that could be obtained by the bisection method (x ≈ -0.82667)

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