#1**0 **

Funny question :)

I assume that log is the natural log and I will use ln as the natural log below.

\(\ln(\ln(i))\\ = \ln\left(\dfrac{1}{2}\cdot\ln(-1)\right)\\ = -\ln 2 + \ln(i\cdot\pi)\\ =-\ln2+\ln i + \ln \pi\\ =\ln\pi-\ln2+\dfrac{1}{2}\ln(-1)\\ =\ln\pi - \ln 2 + \dfrac{\pi}{2}\cdot i\)

I am going to assume that log is log base 10 below and do it again :)

\(\quad\log_{10}\left(\log_{10}(i)\right)\\ =\dfrac{\ln(\frac{\ln i}{\ln10})}{\ln 10}\\ =\dfrac{\ln\left(\ln(i)\right)-\ln\left(\ln 10\right)}{\ln 10}\\ =\dfrac{\ln(i\cdot \pi)-\ln(\ln 10)}{\ln10}\\ =\dfrac{i\cdot \pi+\ln \pi-\ln(\ln10)}{\ln10}\)

MaxWong
Jun 3, 2017