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lnX = 1- ln(x+2) I've done this so far: (x/x+2)=1 e^2=(x/x+2) e^2(x+2)=x Idk if I am starting this right. Our teacher gave us an example problem similar to this and I am just following his steps. I feel like it's wayy off though. How would I get the correct answer? Answer: -1+sqrt(1+e) any help is appreciated, please and thank you!

Precalculus
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why did you divide? ((x/x+2)=1) and you forgot a ln . isn't it multiply? (ln(x(x+2) = 1) but your math is right.
I just saw this and I subtracted instead of multiplying. But I am not sure what to do after this. because my "1" turns into an "e^1" right?
*i mean i subtracted (x+2) in stead of adding to both sides

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Other answers:

yeah. the subtract/add was the main mistake
ok. My main problem is the "e" now. I'm not sure where I place it from here on and how I would get it in my answer :O
I've done this so far now: ln(x(x+2) = 1 x^2+2x=e^1 ??? It's just the e that Im getting stuck on
@Ikushinkan You teacher did NOT write (x/x+2). Notice how this is just 3 unless x = 0. Don't change variable names mid-equation ln(x) = 1- ln(x+2) Observe. Before you do anything else, notice that x > 0. Do you see why? Rearrange Using Algebra Rules ln(x) + ln(x+2) = 1 Use Logarithm Rules ln(x(x+2)) = 1 Use Logarithm Rules \(x(x+2) = e^{1} = e\) Rearrange Using Algebra Rules \(x^{2} + 2x - e = 0\) Solve, using the Quadratic Formula. Don't try to factor that. Remember, throw out the Negative result. Do you know why?
treat "e" as a number. you almost got it. move e to the other side use quadratic formula yeah @tkhunny explained it well. :) i don't think you ALWAYS throw out the negative result, i think. i think only throw it out if you plug it back in to the original equation and it doesn't work, then throw it out.
No, in this case, it is clear and obvious that negative values for x are not appropriate. You do not ALWAYS discard them, only when it is appropriate to do so. The Domain of the logarithm function, y = log(x) is x > 0. No matter what you do after youobserve that, the Domain of the problem is still at least x > 0. There is no need to try negative values in the original problem statement. Simply discard them because they are not in the Domain. You may try them in the original equatin if you like, but it is a waste of time for something you should have seen from the beginning.
no i know, but i was talking about in general. sorry for the confusion.
Good call. That's why I put it under "Observe". It is hoped that other things will be observed for other problems. :-)
ok I have: -1 +-sqrt(2^2 - 4(1)(e))/2(1) -1+-sqrt(4-4e)/2 but how does the two on the bottom cancel out??? D:
i don't think it does you could just plug it into a calculator...?
Application Error. \(x^{2} +2x−e=0\) You started with b = 2. That happened inside the radical , but something else happened at the beginning. Also, c = -e, not +e. Notation Error. -2 +-sqrt(2^2 - 4(1)(-e))/2(1) This is no good. There are parentheses missing. Should Be: [-2 +/- sqrt(2^2 - 4(1)(-e))]/2(1). The parentheses (big square brackets in this case) are NOT optional.
Ok so Im getting this answer now: : [-2 +/- sqrt(2^2 - 4(1)(-e))]/2(1) how does that turn into -1+ sqrt(1+e) ? It isnt making sense to me
What isn't making sense? 1) Why didn't you get two results? 2) e is just a number. If it has said sqrt(1+4), woudl that be a problem? sqrt(1+e) is just fine. You may just need to get used to it.
this is what you should have @Ikushinkan \[x = \frac{ -1 \pm \sqrt{(2)^2-4(1)(-e)} }{ 2(1) } \] maybe that'll make more sense? i didn't do anything; i just wrote it in the equation editor. remember e is a constant e = 2.718281828
Yep! That is exactly what I got. I'm just not sure how that turns into "sqrt(1+e)". where did the "1" inside the aquare root come from?
*square root
it's the quadratic equation... \[x = \frac{ -b \pm \sqrt {b^2-4ac} }{ 2a }\]
@Ikushinkan did you get it?
Yes I did! Thanks for your help!:)
cool ^_^ no problem :)

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