User:ShohagS

• 
  Abdullah AL Shohag
• 
  Abdullah AL Shohag is with his friend
• 
  Abdullah AL Shohag
• 
  Abdullah AL Shohag

"An equation for me has no meaning, unless it expresses a thought of God" - S. Ramanujan

${\displaystyle \log(-x)=y}$

${\displaystyle \Longrightarrow -x=10^{y}}$

${\displaystyle \Longrightarrow -x=10^{a+ib}}$${\displaystyle \Longrightarrow -x=10^{a}.10^{ib}}$

${\displaystyle \Longrightarrow -x=10^{a}.e^{ib.\ln 10}}$

${\displaystyle \Longrightarrow -x=10^{a}.{\biggl (}\cos b+i\sin b{\biggr )}^{\ln 10}}$

If we consider ${\displaystyle i\sin b=0}$, then ${\displaystyle b=n\pi }$ but ${\displaystyle \cos b}$ will be negative only when ${\displaystyle b=(2n-1)\pi }$ where ${\displaystyle n\in \mathbb {N} }$ (for simplicity we will take ${\displaystyle n=1}$, so ${\displaystyle b=\pi }$)

${\displaystyle \Longrightarrow -x=10^{a}.(-1)^{\ln 10}}$

${\displaystyle \Longrightarrow \log(-x)=a+{\ln 10}.log(-1)}$

${\displaystyle \Longrightarrow {\log e}.{\ln(-x)}=a+{\ln 10}.{\log e}.{i\pi }}$ /* ${\displaystyle \log(-1)=i\pi \log e}$ */

${\displaystyle \Longrightarrow {\log e}.{\biggl (}{\ln(x)}+{i\pi }{\biggr )}=a+{\ln 10}.{\log e}.{i\pi }}$

${\displaystyle \Longrightarrow a={\log e}.{\ln x}+{i\pi }{\biggl (}1-{\ln 10}{\biggr )}{\log e}}$

if we put the value of ${\displaystyle a}$ in the marked line,

${\displaystyle \Longrightarrow \log(-x)={\log e}.{\ln x}+{i\pi }{\biggl (}1-{\ln 10}{\biggr )}{\log e}+{\ln 10}.log(-1)}$

${\displaystyle \Longrightarrow \log(-x)={\log x}+{i\pi }{\log e}}$

so if we generalize it for any base, we get,

${\displaystyle \log _{a}(-x)={\log _{a}x}+{i\pi }{\log _{a}e}}$, where ${\displaystyle a>1}$