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

2 Answers
Jul 15, 2016

Answer:

#x=-4#

Explanation:

#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#

Jul 15, 2016

Answer:

#x = -4#

Explanation:

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#