SOLUTION: Solve
ln(4x+3)=ln(7x+9)
log5x+log(x-1)=2
Algebra.Com
Question 183593: Solve
ln(4x+3)=ln(7x+9)
log5x+log(x-1)=2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve
ln(4x+3)=ln(7x+9)
4x+3 = 7x+9
3x = -6
x = -2
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Checking those values in the original equations you get:
ln(-8+3) = ln(-14+9)
ln(-5) = ln(-5)
That looks good but it isn't.
There are no natural logs of negative numbers.
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Your problem has no solution.
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log5x+log(x-1)=2
log[5x(x-1)] = 2
5x^2-5x = 10^2
5x^2 - 5x - 100 = 0
x^2 -x - 20 = 0
(x-5)(x+4) = 0
x = 5 or x = -4
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x = 5 is a good solution; x = -4 just doesn't work in the original equation.
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Cheers,
Stan H.-
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