How to expend $\log_a(\log_ax)$ for $a\in(0;1) \land x\in(0;1)$? Here are some logarithm rules : * $\log_ay=\frac{\ln y}{\ln a}$ * $\log_a(A\cdot B)=\log_aA+\log_aB$ * $\frac{1}{\ln a}=\log_ae$ Hence: $$\log_a(\log_ax)=\log_a \left(\frac{1}{\ln a}\cdot \ln x\right) = \log_a(\log_ae)+\log_a(\ln x) $$ **The problem** : $\ln a<0 \vee \log_ae<0$ to choose- the argument of $\log_a$ is negative. $\ln x$ as well. * * * Origin of the question: $\log_ax=a^x$ =x$ I am not sure if I can say: $\log_ax=\frac{|\ln x|}{|\ln a|}$

Since $\ln a$ and $\ln x$ are negative- we can solve via the complex logarithm: $$ \ln(r\cdot e^{i\theta})=\ln r+i\theta $$ Trick: $e^{i\pi}=-1$

Hence: $$ \ln(-r)= \ln r+i\pi $$ For better calculating: $$ \ln(y)\rightarrow \ln(|y|)+i\pi ;y<0 $$ This way: $$\log_a(\log_ax)=\frac{1}{\ln a}\cdot \ln\left(\frac{\ln x}{\ln a}\right) = \frac{1}{\ln a} \cdot [\ln(\ln x)-\ln(\ln a)] $$

Note that:$$ \ln(\ln x)-\ln(\ln a)=\ln(|\ln x|)+i\pi-(\ln(|\ln a|)+i\pi)=\ln(|\ln x|)-\ln(|\ln a|) $$

Hence: $$\log_a(\log_ax)=\frac{1}{\ln a}\cdot [\ln(|\ln x|)-\ln(|\ln a|)] $$