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Ikushinkan Group Title

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!

  • one year ago
  • one year ago

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  1. jennychan12 Group Title
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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.

    • one year ago
  2. Ikushinkan Group Title
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    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?

    • one year ago
  3. Ikushinkan Group Title
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    *i mean i subtracted (x+2) in stead of adding to both sides

    • one year ago
  4. jennychan12 Group Title
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    yeah. the subtract/add was the main mistake

    • one year ago
  5. Ikushinkan Group Title
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    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

    • one year ago
  6. Ikushinkan Group Title
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    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

    • one year ago
  7. tkhunny Group Title
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    @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?

    • one year ago
  8. jennychan12 Group Title
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    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.

    • one year ago
  9. tkhunny Group Title
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    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.

    • one year ago
  10. jennychan12 Group Title
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    no i know, but i was talking about in general. sorry for the confusion.

    • one year ago
  11. tkhunny Group Title
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    Good call. That's why I put it under "Observe". It is hoped that other things will be observed for other problems. :-)

    • one year ago
  12. Ikushinkan Group Title
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    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:

    • one year ago
  13. jennychan12 Group Title
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    i don't think it does you could just plug it into a calculator...?

    • one year ago
  14. tkhunny Group Title
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    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.

    • one year ago
  15. Ikushinkan Group Title
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    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

    • one year ago
  16. tkhunny Group Title
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    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.

    • one year ago
  17. jennychan12 Group Title
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    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

    • one year ago
  18. Ikushinkan Group Title
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    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?

    • one year ago
  19. Ikushinkan Group Title
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    *square root

    • one year ago
  20. jennychan12 Group Title
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    it's the quadratic equation... \[x = \frac{ -b \pm \sqrt {b^2-4ac} }{ 2a }\]

    • one year ago
  21. jennychan12 Group Title
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    @Ikushinkan did you get it?

    • one year ago
  22. Ikushinkan Group Title
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    Yes I did! Thanks for your help!:)

    • one year ago
  23. jennychan12 Group Title
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    cool ^_^ no problem :)

    • one year ago
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