Golden Ratio In A Logarithmic Equation?

By the change of base rule, we can write the given equation as

$\dfrac{1}{\log_{b}(a)} = \log_{b}(a) + 1 \Longrightarrow (\log_{b}(a))^{2} + \log_{b}(a) - 1 = 0$ , and so $\log_{b}(a) = \dfrac{-1 \pm \sqrt{5}}{2}.$

With $\phi = \dfrac{\sqrt{5} + 1}{2}$ this implies that either $\log_{b}(a) = \phi - 1$ or $\log_{b}(a) = -\phi,$

i.e., that either $\boxed{a = b^{\phi - 1}}$ or $a = \boxed{\dfrac{1}{b^{\phi}}}.$