Hi,
kannst ja einen Basiswechsel durchführen. Allgemein gilt hier:
$$\log_{a}(b) = \frac{\log(b)}{\log(a)}$$
Bei uns also:
$$\log_{a}(x) + \log_{\frac 1a}(x) = \frac{\log(x)}{\log(a)} + \frac{\log(x)}{\log(\frac1a)} = \frac{\log(x)}{\log(a)} + \frac{\log(x)}{\log(1)-\log(a)}$$
$$\frac{\log(x)}{\log(a)} + \frac{\log(x)}{-\log(a)} = \frac{\log(x)}{\log(a)} - \frac{\log(x)}{\log(a)} = 0$$
Alles klar?
Grüße