User:ShohagS

Abdullah AL Shohag

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

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

Defining negative value of logarithms

$y=\log(-x)$

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

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

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

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

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

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

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

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

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

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

if we put the value of $a$ in the marked line,

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

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

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

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