Natural Logarithms

[PolyFX]

1. The problem statement, all variables and given/known data

lnx+ ln(x-1) = 1

solve each equation for x

2. Relevant equations

ln(e^x) = x
e^lnx = x

3. The attempt at a solution

x + (x-1) = e^1 [==> using ln(e^x) = x]
from this point on, I am stuck because I am having trouble isolating x because of the x that is in the brackets.

-Thanks in advance

[Bohrok]

Remember the logarithm property
ln(a) + ln(b) = ln(ab)

[PolyFX]

Bohrok,

Thank you. For some reason I always assumed that the logarithm laws such as log(x/y)= log x-logy etc, could not be applied to natural logarithms. I guess questions like that never really came up.

So using ln(ab) = lna + lnb

I get,

lnx + ln(x-1) = 1
ln(x2-x) = 1

so e1 = x2-x

Am I missing a step? I still cannot isolate x to solve for it. I think I may be jumping in a little too early for the cancellation rule.

-Thanks in advance

[Bohrok]

The properties of logarithms work with any positive number base (I believe), at least with most numbers you come across, like 10 and e.

What you have now is a quadratic equation with an x2 term, so use the quadratic formula after you set the equation equal to 0.