How do you solve #log_3 (5x+5) - log_3 (x^2 - 1) = 0#?

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Answer 1 · 2015-05-11T23:14:32

The answer is #x=6#

I recall the rule:
#log(a)-log(b)=log(a/b)#
and #log(a)=0 <=> a=1#

so #log_3((5x+5)/(x^2-1))=0 <=>#

#<=> (5x+5)/(x^2-1)=1 <=> 5x+5=x^2-1#

So we gotta solve #x^2-5x-6=0#

#(5+-sqrt(25+24))/2 = (5+-7)/2 => x=6 or x=-1#

6 is an acceptable solution:

#5*6+5=35>0# and #36-1=35>0#

-1 is not an acceptable answer:
#-5+5=0 and log(0)# is not defined, so that solution is an extraneous solution.