help5

what are you asking?

a^x=x^y

a^y=x^x

$\begin{array}{rcll} (1) & a^x &=& x^y \qquad & | \qquad \ln{()} \\ & x\cdot \ln{(a)} &=& y \cdot \ln{(x)} \\ & y &=& \frac{x\cdot \ln{(a)} }{ \ln{(x)} } \\ \hline \\ (2) & a^y &=& x^x \qquad & | \qquad \ln{()} \\ & y\cdot \ln{(a)} &=& x \cdot \ln{(x)} \qquad & | \qquad y = \frac{x\cdot \ln{(a)} }{ \ln{(x)} }\\\\ & \frac{x\cdot \ln{(a)} }{ \ln{(x)} }\cdot \ln{(a)} &=& x \cdot \ln{(x)} \\ & x\cdot [\ln{(a)}]^2 &=& x \cdot [\ln{(x)}]^2 \\ & x\cdot [\ln{(a)}]^2 - x \cdot [\ln{(x)}]^2 &=& 0 \\ & \underbrace{x}_{=0} \cdot \underbrace{ \{~ [\ln{(a)}]^2 - [\ln{(x)}]^2 ~\} }_{=0} &=& 0 \\\\ & \mathbf{x_1} & \mathbf{=} & \mathbf{0} \\\\ & \{~ [\ln{(a)}]^2 - [\ln{(x)}]^2 ~\} &=& 0 \\ & [\ln{(a)}]^2 - [\ln{(x)}]^2 &=& 0 \\ & [\ln{(a)}]^2 &=& [\ln{(x)}]^2 \\ & \ln{(a)} &=& \ln{(x)} \\ & a &=& x \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{a} \\ \hline \\ & y &=& \frac{x\cdot \ln{(a)} }{ \ln{(x)} } \\ & y &=& \frac{a\cdot \ln{(a)} }{ \ln{(a)} } \\ & \mathbf{y} &\mathbf{=}& \mathbf{a} \\ \end{array}$