How do you solve #log_b(x^2 -2) + 2log_b 6 = log_b6x#?

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Answer 1 · 2016-05-21T10:22:47

#x={3/2,4/3}#

#log_b(x^2-2)+2log_b6=log_b6x#

#log_b(x^2-2)+log_b6^2=log_b6x#

#log_b(x^2-2)*6^2=log_b6x#

#log_b(x^2-2)*6^2-log_b6x=0#

#log_b(((x^2-2)*6^2)/(6x))=0#

#log_b((6x^2-12)/x)=0#

#(6x^2-12)/x=b^0#

#b^0=1#

#(6x^2-12)/x=1#

#6x^2-12=x" "6x^2-x-12=0#

#(2x-3)(3x+4)=0#

#2x-3=0" "2x=3" " x=3/2#

#3x+4=0" "3x=4" "x=4/3#

#x={3/2,4/3}#