Question

Evaluate log(-13) Correct answer will get brainliest

Answer 1

if given log(a)=b, where no base is stated, we assume base 10, or $log(a)=log_{10}(a)$

also, $log_a(b)=c$ translates to/can be written equivilently as $a^c=b$

therefore, given

$log(-13)=?$ where we want to find ?

assume base 10 and translate

$10^?=-13$

since we cannot raise a positive number to any real power and give a negative number, ? is not a real number and therefore, there are no real solutions

if we do want to find a solution, we can use Euler's idendity

$e^{\pi i}=-1$ where i is the complex number i=√-1 and e is euler's number

we can try to change bases to find the value of ?

$e^{\pi i}=-1$

$13e^{\pi i}=-13$

so

$13e^{\pi i}=10^?$

taking ln of both sides

$ln(13e^{\pi i})=ln(10^?)$

using log rules

$ln(13)+(\pi i)ln(e)=(?)ln(10)$

$ln(13)+\pi i=?ln(10)$

divie both side by ln(10)

$\frac{ln(13)+\pi i}{ln(10)}=?$

in a+bi form

$?=\frac{ln(13)}{ln(10)}+\frac{\pi}{ln(10)}i$

there are no real numbers that it evaluates to

however, it does evaluate to a complex number which is $\frac{ln(13)}{ln(10)}+\frac{\pi}{ln(10)}i$ or aproximately 1.11394+1.36438i