Asked by Bob

Need help solving for x with work shown

2log6(4x)=0

Log2(x)+log2(x-3)=2

Answers

Answered by Damon
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 :)
Answered by Damon
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

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