Can someone please help solve inverting an equation?

I have this formula that works well to convert wire gauge to mm:

mm dia = 0.127 * 92 ^ ((36-AWG)/39)

Following this, the same source offers the following as an inverse formula(mm to gauge):

AWG = -(ln (dia. mm/0.127) / (ln (92) * 39) – 36)

$\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \qquad \text{AWG} =\ ? \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad : 0.127 \\\\ \dfrac{ dia_{mm} }{0.127} &=& 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad \ln{()}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \ln{ \left( 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \right)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \left( \frac{36-\text{AWG}}{39} \right) \cdot \ln{(92)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& ( 36-\text{AWG} ) \cdot \frac{ \ln{(92)} } {39} \quad &|\quad \cdot \frac{39}{ \ln{(92)} } \\\\ \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 - \text{AWG} \quad &|\quad +\text{AWG} \\\\ \text{AWG} + \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 \quad &|\quad -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} \\\\ \mathbf{\text{AWG} } &\mathbf{=}& \mathbf{36 -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} }\\ \hline \end{array}$