I was in the shower and I was just thinking is the solution to $\displaystyle 1-\log_a(x)=-\log_x(a)$
$\displaystyle x=a^{\phi}$?
Well, I find the same thing yep ~
I can't work with $\displaystyle \log_a$ :
$\displaystyle 1-\frac{\ln(x)}{\ln(a)}=-\frac{\ln(a)}{\ln(x)}$
$\displaystyle \ln(a)\ln(x)-\ln^2(x)=-\ln^2(a)$
Dividing by $\displaystyle \ln^2(a)$ :
$\displaystyle (\frac{\ln(x)}{\ln(a)})^2-\frac{\ln(x)}{\ln(a)}-1=0$
Hence $\displaystyle x=a^\varphi$
Is it what you were thinking about ?
PS : I rather prefer $\displaystyle \varphi$ than $\displaystyle \phi$
Isnt that cool...I did it a different way $\displaystyle 1-\log_a(x)=-\log_x(a)\Rightarrow{1-\log_a(x)=\frac{-1}{log_a(x)}}$...then letting $\displaystyle \log_a(x)=u$ and multiplying through and using quadratic formula I came up with $\displaystyle \phi$...thats right $\displaystyle \phi$