How to get the graph for $y= \log_{1/a} (x)$ from $y= \log_a (x)$?

Question

I know that it is symmetric the Ox axle, but I can't prove it.

Answer 1

If $y=\log_a x$ then $x=a^y=\left(\frac 1a\right)^{-y}$. Therefore $\log_{\frac 1a} x=-y$

I always find it easier to write these log equations in terms of exponentials if I am not quite sure what is going on.

Answer 2

I think there is a property of logarithm that you need. $$y = \log_{1/a}(x) = -\log_a (x)$$