How do you solve #log_2((3x-2)/5)=log_2(8/x)#?

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Answer 1 · 2018-05-10T13:14:48

#x=4#

#log_"2"((3x-2)/5)=log_"2"(8/x)#

#D_"f":x in RR_"+"^"*"#

#(3x-2)/5=8/x#

#3x²-2x=40#

#3x²-2x-40=0#

#Δ=4+4*3*40#

#Δ=484#

#x_"1"=(2-sqrt(484))/6#

#x_"2"=(2+sqrt(484))/6#

#cancel(x_"1"=-10/3)# impossible due to our #D_"f"#

#x_"2"=4#

\0/ here's our answer!

Answer 2 · 2018-07-05T02:50:53

#color(maroon)(x = 4#

#log_2 ((3x-2)/5) = log_2 (8/x)#

#(3x-2)/5 = (8/x), " Removing Log on both sides"#

#3x^2 - 2x = 40#

#3x^2 - 12x + 10x - 40 = 0#

#3x (x - 4) + 10 (x - 4) = 0#

#(x - 4) * (3x + 10) = 0#

#x = 4, color(crimson)(cancel(-10/3)#

#color(maroon)(x = 4#