2log6(4x)=0
log6 (16 x^2) = 0
6^log6 (16 x^2) = 16x^2 = 6^0 = 1
so
x^2 = 1/16
x = 1/4
normally I would say also -1/4 but try to take the log of -1 :)
Need help solving for x with work shown
2log6(4x)=0
Log2(x)+log2(x-3)=2
2 answers
Log2(x)+log2(x-3)=2
2^[Log2(x)+log2(x-3)] =2^2
2^log2(x^2-3x) = 2^2
(x^2-3x) = 4
x^2 - 3 x - 4 = 0
(x-4)(x+1) = 0
x = 4 or x = -1
2^[Log2(x)+log2(x-3)] =2^2
2^log2(x^2-3x) = 2^2
(x^2-3x) = 4
x^2 - 3 x - 4 = 0
(x-4)(x+1) = 0
x = 4 or x = -1
1 answer