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

2 Answers
Aug 14, 2018

x = 7

Explanation:

log_b m = log_b n => m = n

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

2x + 1 = 3x - 6

x = 7

Aug 14, 2018

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

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

Explanation:

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