SOLUTION: log2=log(6+4x)+log(3x+6)
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Question 696404: log2=log(6+4x)+log(3x+6)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log2=log(6+4x)+log(3x+6)
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log[(4x+6)(3x+6)] = log(2)
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(4x+6)(3x+6) = 2
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2*3(x+3)(x+2) = 2
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(x+3)(x+2) = 1/3
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x^2 + 5x + 6 = 1/3
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3x^2 + 15x + 17 = 0
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Solve using the Quadratic Formula.
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x = -1.7362
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Note: That value of "x" makes 4x+6 be negative.
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Conclusion: No solution.
Cheers,
Stan H.
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