Question

solve: lnx + ln(x-1) = 1

Answer 1

\[\Large \ln x+\ln(x-1)=1\]
First we want to apply this rule of logarithms:\[\large \color{royalblue}{\log(a)+\log(b)=\log\left(a\cdot b\right)}\]

Answer 2

okay

Answer 3

so it would be ln(x^2+x)

Answer 4

\[\Large \ln(x^2+x)=1\]
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.

Answer 5

so it would be \[e ^1 = x^2 +x\]

Answer 6

Yes good! :)

Answer 7

but then I have to solve for x which confuses me

Answer 8

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,\[\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 9

wouldnt it be like a weird answer, like (x-ex)(x+ex) or something like that

Answer 10

Ya it's definitely going to be a weird answer :) It won't factor nicely so we have to use the quadratic formula.

Answer 11

\[\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \quad\to\quad x=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-e)}}{2\cdot1}\]
Understand how I plugged those in? :o

Answer 12

yup

Answer 13

so it would be (-1 +- root3e)/2 ?

Answer 14

I think it's going to be, \[\Large x=\frac{-1\pm\sqrt{1+4e}}{2}\]

Answer 15

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.\[\large x=\frac{-1-\sqrt{1+4e}}{2}\quad=\quad?\]\[\large x=\frac{-1+\sqrt{1+4e}}{2}\quad=\quad?\]

Answer 16

approx -2.22 and 1.22

Answer 17

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.\[\Large \ln(-2.22) \quad=\quad ?\]

Answer 18

it says error on my calculator :p

Answer 19

Interesting! :O

Answer 20

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.

Answer 21

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 :)

Answer 22

Woops I typoed,\[\Large x=\frac{-1+\sqrt{1+4e}}{2}\]

Answer 23

so there is one x answer and that will be my domain? so it's (-infinity,1)u(1.22,infinity) ?

Answer 24

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?

Answer 25

no, i didn't need a domain. I thought I did :p
thanks so much! you're a life saver!

Answer 26

yay team \c:/