How do you solve #\log _ { 2} - m = \log _ { 2} 5m#?

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Answer 1 · 2018-07-31T12:57:07

I found no Real solutions;

I am not sure but I think the idea is to solve:

#log_2(-m)=log_2(5m)#

let us take the exponetial of #2# of both sides:

#2^(log_2(-m))=2^(log_2(5m))#

this allows us to get rid of the logs and write:

#cancel(2)^(cancel(log_2)(-m))=cancel(2)^(cancel(log_2)(5m))#

#-m=5m#

#6m=0#

#m=0#

that cannot be accepted because we cannot evaluate #log_2(0)#. So our equation has no Real solutions.