How do you evaluate ${log}_{144} (- 12)$ $log_144 (-12)$ ?

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Answer 1 · 2016-11-19T05:25:44

Though unreal, the answer in complex form is

$\frac{1}{2} + i \frac{(4 k + 1) \frac{π}{2}}{ln 12}$ $1/2 + i((4k+1)pi/2)/ln 12$ , k = 0, +-1, +-2, +-3, ... .

#log_144 (-12) is unreal. Yet, I evaluate it for you.

Use $i^{2} = - 1$ $i^2=-1$ .

${log}_{144} (- 12)$ $log_144(-12)$

$= \frac{ln (- 12)}{ln 144}$ $=ln (-12)/ln 144$

$= \frac{ln (i^{2} 12)}{{ln 12}^{2}}$ $=ln(i^2 12)/ln 12^2$

$= \frac{2 ln i + ln 12}{2 ln 12}$ $=(2 ln i + ln 12)/(2 ln 12)$

$\frac{ln i}{ln 12} + \frac{1}{2}$ $ln i/ln12+1/2$

Interestingly,

i = e^(i(4k+1)pi/2), k = 0, +-1, +-2, +-3, ...#

So, $ln i = ln (e (i (4 k + 1) \frac{π}{2}))$ $ln i = ln (e(i(4k+1)pi/2))$

$= i (4 k + 1) \frac{π}{2}$ $=i(4k+1)pi/2$ .

So, the answer is

$\frac{1}{2} + i \frac{(4 k + 1) \frac{π}{2}}{ln 12}$ $1/2 + i((4k+1)pi/2)/ln 12$ , k = 0, +-1, +-2, +-3, ...#

How do you evaluate log144(−12)log_144 (-12)?