anonymous
  • anonymous
solve: lnx + ln(x-1) = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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zepdrix
  • zepdrix
\[\Large \ln x+\ln(x-1)=1\] First we want to apply this rule of logarithms:\[\large \color{royalblue}{\log(a)+\log(b)=\log\left(a\cdot b\right)}\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
so it would be ln(x^2+x)

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zepdrix
  • zepdrix
\[\Large \ln(x^2+x)=1\] Mmm ok looks good. So for the next step, we can either exponentiate each side, which will deal with the logarithm. Or, if you remember how to rewrite a log in exponent form we can skip that step.
anonymous
  • anonymous
so it would be \[e ^1 = x^2 +x\]
zepdrix
  • zepdrix
Yes good! :)
anonymous
  • anonymous
but then I have to solve for x which confuses me
zepdrix
  • zepdrix
Hmm I guess we could start by subtracting e from each side,\[\Large 0=x^2+x-e\]And next let's throw this into the Quadratic Formula to find solutions for x,\[\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
wouldnt it be like a weird answer, like (x-ex)(x+ex) or something like that
zepdrix
  • zepdrix
Ya it's definitely going to be a weird answer :) It won't factor nicely so we have to use the quadratic formula.
zepdrix
  • zepdrix
\[\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \quad\to\quad x=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-e)}}{2\cdot1}\] Understand how I plugged those in? :o
anonymous
  • anonymous
yup
anonymous
  • anonymous
so it would be (-1 +- root3e)/2 ?
zepdrix
  • zepdrix
I think it's going to be, \[\Large x=\frac{-1\pm\sqrt{1+4e}}{2}\]
zepdrix
  • zepdrix
But we'll need to do a little bit of extra work. We have two possible solutions for x. We need to see if both of them will work for our `original function`. Let's punch these into the calculator, it will be a lot easier to work with that way.\[\large x=\frac{-1-\sqrt{1+4e}}{2}\quad=\quad?\]\[\large x=\frac{-1+\sqrt{1+4e}}{2}\quad=\quad?\]
anonymous
  • anonymous
approx -2.22 and 1.22
zepdrix
  • zepdrix
Mmmm ok good good, that looks right. Will either of those x values give us a problem if we try to plug it into our function?\[\Large \ln x+\ln(x-1)=1\]If you're not sure, try putting them in a see what happens.\[\Large \ln(-2.22) \quad=\quad ?\]
anonymous
  • anonymous
it says error on my calculator :p
zepdrix
  • zepdrix
Interesting! :O
zepdrix
  • zepdrix
Ya we can't plug negative numbers into a log :) Try to remember that!! The domain of ln(x) is \(\large (0,\infty)\). So we can't use the x=-2.22 solution.
zepdrix
  • zepdrix
So our final answer would be,\[\Large x=1.22 \qquad\text{or}\qquad x=\frac{1+\sqrt{1+4e}}{2}\] Depending on whether you prefer to leave it exact or decimal form :)
zepdrix
  • zepdrix
Woops I typoed,\[\Large x=\frac{-1+\sqrt{1+4e}}{2}\]
anonymous
  • anonymous
so there is one x answer and that will be my domain? so it's (-infinity,1)u(1.22,infinity) ?
zepdrix
  • zepdrix
There is one x value that solves the problem. That's the value we found. Do you also need to specific the domain of the function?
anonymous
  • anonymous
no, i didn't need a domain. I thought I did :p thanks so much! you're a life saver!
zepdrix
  • zepdrix
yay team \c:/

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