How do you solve #log_2(3x-4) +log_2(5x-2)=4#?

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Answer 1 · 2016-05-29T10:27:15

#x={-4/15,2}#

#log_a b+log_a c=log_a (b*c)#

#log_2(3x-4)+log_2(5x-2)=4#

#log_2(3x-4)(5x-2)=4#

#log_a b=x" ; "b=a^x#

#(3x-4)(5x-2)=2^4#

#15x^2-6x-20x+8-16=0#

#15x^2-26x-8=0#

#(x-2)(15x+4)=0#

#(x-2)=0" ; " rarr" "x=2#

#(15x+4)=0" ; "rarr " "x=-4/15#

#x={-4/15,2}#