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

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Answer 1 · 2016-05-29T10:27:15

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

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

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

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

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

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

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

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

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

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

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

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

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