How do you solve #\log_{3} ( 2x + 1) = \log_{3} ( 3x - 6)#?

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Answer 1 · 2018-08-14T19:28:18

#x = 7#

#log_b m = log_b n => m = n#

# log_{3} ( 2x + 1) = log_{3} ( 3x - 6)#

# 2x + 1 = 3x - 6#

#x = 7#

Answer 2 · 2018-08-14T19:31:28

#log_3(2x+1)=log_3(3x-6)=>2x+1=3x-6#

#=>-x=-7=>x=7#

Here,

#log_3(2x+1)=log_3(3x-6)#

#log_3(2x+1)-log_3(3x-6)=0#

#log_3((2x+1)/(3x-6))=0to[becauselog_aM-log_aN=log_a(M/N)]#

#:.(2x+1)/(3x-6)=3^0=1#

#:.2x+1=3x-6#

#:.2x-3x=-6-1#

#:.-x=-7#

#:.x=7#