Well this is an interesting question
When you take a logarithm: #log_10 (100) = a # this is like asking what is the value of #a# in #10^a =100 #, or what do you raise 10 to, to get 100
And we know that #a^b# can never be negative...
#y = e^x :# graph{e^x [-10, 10, -5, 5]}
We can see this is never negative, so hence #a^b < 0 # has no solutions
So #log(-100) # is like asking what value for #a# in #10^a = -100 # but we know #10^a# can never be negative, hence no real solution
But what if we wanted to find #log(-100) # using complex numbers...
Shown below
let # omega = log(-100) # ( where #logx -= log_10 x # )
#=> 10^omega = -100 #
#=> e^( omega log_e 10 ) = 100*e^(pi i)* e^(2kpi i ) #
As we know #e^(2kpi i ) = 1 , AA k in ZZ #
#=> e^(omega log_e 10 ) = 100 e^( pi i( 1 + 2k) ) #
#=> omega * log_e 10 = log_e (100e^ ( pi i ( 1+2k)) ) #
#omega*log_e 10 = log_e 100 + pi i (1+2k) #
#color(red)(=> log_10(-100 ) = 1/log_e 10 ( log_e 100 + pi i (1+2k) ) #
# AA k in ZZ # - For all k, that are integers...
