How do you solve #ln8-ln(x+4)=1#?

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Answer 1 · 2016-08-21T17:40:59

#x=8/e-4=-1.057#

#ln8-ln(x+4)=1# can be written as

#ln8-ln(x+4)=lne# or

#ln8/(x+4)=lne# or

#8/(x+4)=e# or

#x+4=8/e#

i.e. #x=8/e-4# or

#x=2.943-4=-1.057#