SOLUTION: please help me solve for x: 2logx=log2+log(3x-4) thanks

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Question 478395: please help me solve for x: 2logx=log2+log(3x-4) thanks
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2logx=log2+log(3x-4)
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2logx=log2+log(3x-4)
2logx-log2-log(3x-4)=0
2logx-(log2+log(3x-4))=0
place under single log
log (x^2/2*(3x-4))=0
convert to exponential form: base(10) raised to log of number(0)=number(x^2/2*(3x-4))
10^0=(x^2/2*(3x-4))
1=x^2/6x-8
x^2=6x-8
x^2-6x+8=0
(x-4)(x-2)=0
x=4
or
x=2