How do you solve for x in #log_6 3x-log_6 (x+1)=log_6 1#?

1 Answer
Gió
Feb 14, 2015

You can use the property of the log:
#log_aM-log_aN=log_a(M/N)#
So you have:
#log_6((3x)/(x+1))=log_6(1)#
#log_6((3x)/(x+1))=0#
Because from the definition od logarithm:
#log_ab=x => a^x=b#
and: #log_6(1)=0#

as for:
#log_6((3x)/(x+1))=0#
again you get:
#((3x)/(x+1))=6^0=1#
#3x=x+1#
#x=1/2#

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