Question 690644: log4(3x-3) = log4(x+1) + 2log4(4)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! log4(3x-3) = log4(x+1) + 2log4(4)
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log4(3x-3) = log4[(x+1)*4^2]
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Equate the inverse logs to solve for "x":
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3x-3 = 16(x+1)
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3x-3 = 16x + 16
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13x = -19
x = -19/13
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Cheers,
Stan H.
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