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Solve the logarithmic equation lnx - ln(x+1) = 2?
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For the beginning, we'll use the quotient property of the logarithms:
lnx - ln(x+1) = ln [x/(x+1)]
Now, we'll have to use the one to one property, that means that:
ln [x/(x+1)] = 2 lne if and only if [x/(x+1)] = e^2
After cross multiplying, we'll get:
x = x*e^2 + e^2
We'll move the terms which are containing the unknown, to the left side:
x - x*e^2 = e^2
After factorizing, we'll get:
x (1-e^2) = e^2
x = e^2/(1-e^2)
But 1-e^2<0, so x = e^2/(1-e^2)<0, which is impossible because x has to be positive!
So, the equation has no solutions.
Posted by giorgiana1976 on May 28, 2010 at 12:36 AM (Answer #1)
High School Teacher
ln x -ln (x+1) = 2
since ln x-ln y = ln x/y
then ln(x/(x+1)= 2
==> e^2 = x/(x+1)
==> (x+1)e^2 = X
==> xe^2 + e^2 = x
==> xe^2 -x= -e^2
==> x(e^2-1) = - e^2
==> x = -e^2/(e^2 -1)
Posted by hala718 on May 28, 2010 at 12:44 AM (Answer #2)
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