How do you solve #ln(-m)=ln(m+10)#?

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Answer 1 · 2016-12-21T00:16:21

#m = -5#

Given:

#ln(-m) = ln(m+10)#

Take exponents of both sides to find:

#-m = e^(ln(-m)) = e^(ln(m+10)) = m+10#

Add #m-10# to both ends to get:

#-10 = 2m#

Divide both sides by #2# and transpose to get:

#m = -5#