User:ShohagS

= Abdullah AL Shohag =



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

= Defining negative value of logarithms =

$$y = \log(-x) $$

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

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

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

$$\Longrightarrow 10^{a}.\biggl(\cos b + i\sin b\biggr)^{\ln10} = -x $$

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

$$\Longrightarrow 10^{a}.(-1)^{\ln10} = -x $$ $$\Longrightarrow a+ {\ln10}.log(-1) = \log(-x) $$    $$\Longrightarrow a + {\ln10}.{\log e} .{ i\pi} = {\log e}.{\ln(-x)} $$                                        /*   $$\log (-1) = i\pi\log e $$  */

$$\Longrightarrow a + {\ln10}.{\log e} .{ i\pi} = {\log e}.\biggl({\ln(x)} + {i\pi}\biggr) $$

$$\Longrightarrow a = {\log e}.{\ln x} + { i\pi}\biggl(1 - {\ln10}\biggr){\log e} $$

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

$$\Longrightarrow {\log e}.{\ln x} + { i\pi}\biggl(1 - {\ln10}\biggr){\log e} + {\ln10}.log(-1) = \log(-x) $$

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

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

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

bn:User:ShohagS