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How do you solve the equation #log_3(x+5)-log_3(x-7)=2#?

1 Answer

Feb 17, 2015

You can use the following facts:

#log_aM-log_aN=log(M/N)# and:

#log_ab=x => a^x=b#

So you get:

#log_3(x+5)-log_3(x-7)=2#
#log_3((x+5)/(x-7))=2#
#(x+5)/(x-7)=3^2#
#x+5=9(x-7)#
#x-9x=-63-5#
#-8x=-68#
#x=68/8=17/2#

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