I am having trouble solving for a in the problem below. 
I've simplified it down to: $e^{14} = ln(e^e \cdot a)$. I'm not really sure where to go from here.
2026-03-26 06:02:56.1774504976
Solving a variable in a matrix equation?
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Observe
$$ e^{14} \;\; =\;\; \ln(e^e a) \;\; =\;\; \ln e^e +\ln a \;\; =\;\; e + \ln a $$
thus $\frac{1}{\ln a} = \frac{1}{e^{14} - e}$. Assuming NOTA stands for "none of the above" that should be your answer.