SOLUTION: Ive totally lost myself with this, so if there is anyone that can help please 2log[4](x)+log[4](3)=log[4](x)-log[4](2) All the 4s are log base 4 Thank you

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Ive totally lost myself with this, so if there is anyone that can help please 2log[4](x)+log[4](3)=log[4](x)-log[4](2) All the 4s are log base 4 Thank you      Log On

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Question 398220: Ive totally lost myself with this, so if there is anyone that can help please

2log[4](x)+log[4](3)=log[4](x)-log[4](2)

All the 4s are log base 4
Thank you
Answer by stanbon(57940) About Me  (Show Source):
You can put this solution on YOUR website!
2log[4](x)+log[4](3)=log[4](x)-log[4](2)
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log4(x^2) + log4(3) = log4[x/2]
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log4(3x^2) = log4(x/2)
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3x^2 = x/2
6x^2 = x
6x^2-x = 0
x(6x-1) = 0
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x cannot be zero, so x = 1/6 is the only answer.
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Let me know if you do not understand all this.
Cheers,
Stan H.
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