SOLUTION: log7(8x)=log7(2x)+log7(3x-5)

Question 1004328: log7(8x)=log7(2x)+log7(3x-5)
Found 2 solutions by stanbon, fractalier:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log7(8x)=log7(2x)+log7(3x-5)
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log7(8x) = log7[2x(3x-5)]
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8x = 6x^2-10x
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6x^2 - 18x = 0
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6x(x-3) = 0
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x = 0 or x = 3
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Acceptable answer:: x = 3
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Cheers,
Stan H.
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
I imagine you wish to solve this...
log7(8x)=log7(2x)+log7(3x-5)
log7(8x)=log7[(2x)(3x-5)]
Now exponentiate 7-to-the
8x = (2x)(3x-5)
8x = 6x^2 - 10x
6x^2 - 18x = 0
x^2 - 3x = 0
x(x-3) = 0
x = 0 or x = 3
You cannot take the log of zero, so x = 3.
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