How do you solve #ln(x + 5) - ln(3) = ln(x - 3)#?

1 Answer
Jul 8, 2015

I found #x=7#

Explanation:

You can use one rule of the logs:
#loga-logb=loga/b#
to get:
#ln((x+5)/3)=ln(x-3)#
take the exponential of both sides:
#e^(ln((x+5)/3))=e^(ln(x-3))#
that gives you (cancelling #ln# and #e#):
#(x+5)/3=x-3#
#x+5=3x-9#
#2x=14#
#x=7#