Question

The value of ${\log }_2(-3)$ is

• A. $\frac{\log 2-i\pi }{\log 3}$
• B. $\frac{\log 3+2i\pi }{\log 2}$
• C. $\frac{\log 3+i\pi }{\log 2}$
  
• D. $\frac{\log 2+i\pi }{\log 3}$

Answer 1

The correct option is C $\frac{\log 3+i\pi }{\log 2}$


${\log }_2(-3)=\frac{{\log }_e(-3)}{{\log }_e\left(2\right)}$
$=\frac{{\log }_e\left(3e^{i\pi }\right)}{{\log }_e\left(2\right)}$
${\log }_2(-3)=\frac{\log 3+i\pi }{\log 2}.$