How do you solve #log_19(-5x-6)=log_19(2-3x)#?

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Answer 1 · 2016-07-15T09:10:58

#x=-4#

#log_19(-5x-6)=log_19(2-3x)#

#hArrlog_19(-5x-6)-log_19(2-3x)=0# or

#log_19((-5x-6)/(2-3x))=log_19(1)# or

#(-5x-6)/(2-3x)=1# or

#-5x-6=2-3x# or

#-5x+3x=2+6# or #-2x=8#

or #x=-8/2=-4#

Answer 2 · 2016-07-15T09:29:31

#x = -4#

The log equation leads directly to a linear equation.

if #logA = logB, " " rArr A = B#

#-5x -6 = 2 - 3x#

#-6-2 =-3x+5x#

#-8 = 2x#

#-4 = x#