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anonymous

  • 5 years ago

Please check and correct my work. Logarithm solve for (x) ln(x-3)+ln(x+4)=1 My Steps: -Law 1: ln[(x-3)(x+4)]=1 -e^1=x^2+x-12...?

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  1. anonymous
    • 5 years ago
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    couldn't you simply leave it in the expanded form and e both sides it would give you x=e-1/2

  2. anonymous
    • 5 years ago
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    if you do it the way you were doing it you have to use the quadratic formula your steps are right

  3. anonymous
    • 5 years ago
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    Okay now im a bit confused. could you show me how

  4. anonymous
    • 5 years ago
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    well e^ln(x-3)+e^ln(x+4)=e^1 which is x-3+x+4=e, 2x+1=e solve for x

  5. anonymous
    • 5 years ago
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    don't listen to calc2student he doesn't know what he is talking about, that is not a rule of logs aka e^(ln(x-3)+ln(x+4) =\ x-3+x+4

  6. anonymous
    • 5 years ago
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    repeat you did it correct do not listen to calc2 student, x=(e-1)/2=0.859.... your method the right one x=-4.37, +3.37

  7. anonymous
    • 5 years ago
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    please this is the second question you have answered incorrectly calc2student stop responding to questions if you don't know how to help

  8. Vijay
    • 5 years ago
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    x = 6.7 and 0.29

  9. Vijay
    • 5 years ago
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    if u need more help, i can explain

  10. anonymous
    • 5 years ago
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    sigh.... (-b+-(b^2-4(a)(c))^1/2)/2 = (-1+-(1-(4*-14.72)^1/2)/2 = (-1+-7.74)/2 = -4.37 or +3.37 feel free to check my work if you need to oh btw x^2+x-12-2.71; a=1, b=1, c=-14.71

  11. anonymous
    • 5 years ago
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    well i put it in and it says its wrong...

  12. Vijay
    • 5 years ago
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    can u make sure is that log or ln ?

  13. Vijay
    • 5 years ago
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    if it is log my answer is correct

  14. anonymous
    • 5 years ago
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    ln(x-3)+ln(x+4)=1 ...so its ln

  15. anonymous
    • 5 years ago
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    it should just be 3.37 b/c you can't take the the natural log of a negative number, check wolfram alpha it concurs with me..... if you have never used www.wolframalpha.com I highly suggest you try it :)

  16. anonymous
    • 5 years ago
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    can u help me with this one?

  17. anonymous
    • 5 years ago
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    solve the equation. Round to nearest 4 decimal place if necessary (x^2)(5^x)-(5^x)=0

  18. anonymous
    • 5 years ago
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    1,-1

  19. anonymous
    • 5 years ago
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    (x^2)(5^x)-(5^x)=0-->5^x(x^2-1)=0-->x^2-1=0-->x=+-(1)^(1/2)-->x=+1,-1

  20. anonymous
    • 5 years ago
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    (x^2)(10^x)-(x)(10^x)=6(10^x)

  21. anonymous
    • 5 years ago
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    and thanks for the steps, greatly appreciated and needed for me to study

  22. anonymous
    • 5 years ago
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    same factor out the 10^x---> x^2-x=6--->x^2-x-6-->x=3,-2

  23. anonymous
    • 5 years ago
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    i have this problem. but it has this weird sign after. i posted it on google documents so you could see it. I thought maybe it would affect the question is some way....heres the link: https://docs.google.com/document/d/1C3mBIu2pp2Q1LUPo4I-ABl6XFxG56PxOrmH6qFAnxJg/edit?hl=en&authkey=CIKk49gI ...if the link works

  24. anonymous
    • 5 years ago
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    now your pushing my algebra skillz to the limit..... so it maybe 1/2 but I'm not sure sorry

  25. anonymous
    • 5 years ago
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    or it could be 8.... lol sorry really I don't really know this one

  26. anonymous
    • 5 years ago
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    lol.... no for real it could be 2.... this time I think its right :)

  27. Vijay
    • 5 years ago
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    ln(x-3)+ln(x+4)=1 lnx / ln 3 + lnx * ln4 = 1 ln(x) [1/ln3 + ln4]=1 ln(x)[1+ln3*ln4]=ln3 ln(x)= ln3/(1+ln(3)*ln(4) = ln3(1+ln7)

  28. Vijay
    • 5 years ago
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    last line is = ln3 / (1+ln7)

  29. anonymous
    • 5 years ago
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    was I wrong?

  30. anonymous
    • 5 years ago
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    how about x=-0.7937

  31. Vijay
    • 5 years ago
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    x should be more than 3, because there is ln(x-3), ln will not take negative value

  32. anonymous
    • 5 years ago
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    uuggghhh.... yeah we've deciphered that he posted a link to another problem.... and btw you were definitely wrong on your last response to that problem.... in fact I have no idea what you were doing @vijay

  33. anonymous
    • 5 years ago
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    vijay, you have your log properties mixed up. ln(a) - ln(b) = ln(a / b), not the other way around. Donovan, your original post was all correct. You should have: (x - 3)(x + 4) = e x^2 + x - 12 = e x^2 + x = e + 12 x^2 + x + 1/4 = e + 12 + 1/4 (x + 1/2)^2 = e + 49/4 \[(x + 1/2)^2=(4e+49)/4\] \[x + 1/2 = \pm \sqrt{4e+49}/2\] \[x = (-1\pm \sqrt{4e+49})/2\] Only the positive version is going to work, which is x = 3.369. The negative version results in a negative x which won't work.

  34. anonymous
    • 5 years ago
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    yeah I posted that like days ago @branlegr check out the link he posted and tell me what you get

  35. anonymous
    • 5 years ago
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    @branlegr I'm a huge fan of wolfram as well ;)

  36. anonymous
    • 5 years ago
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    Haha, I just saw it. Got kinda mixed up in all this back-and-forth of incorrect methods.

  37. anonymous
    • 5 years ago
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    And that problem at that link is a b****.

  38. anonymous
    • 5 years ago
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    yea I'm thinking 2^(1/log3(x))=2^-2-->1/log3(x)=-2--->1=-2log3(x)-->-1/2=log3(x)--->x^3=-1/2--->x=(-1/2)^(1/3)=~-0.7937.... I really do think that one is correct but I'm not sure... oh well

  39. anonymous
    • 5 years ago
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    Imma working on it...

  40. anonymous
    • 5 years ago
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    Answer is 1/sqrt(3). \[2^{1/\log_{3}x }=1/4\] \[\ln 2^{1/\log_{3}x }=\ln 1/4\] \[(1/\log_{3}x)\ln 2=\ln 1/4 \] \[1/\log_{3} x=\ln(1/4)/\ln 2\] \[1 /\log_{3}x=\log_{2}1/4 \] \[1/\log_{3} x=-2\] \[\log_{3} x=-1/2\] \[3 ^{-1/2}=1/\sqrt{3}\]

  41. anonymous
    • 5 years ago
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    nice I was totally wrong.... confused the last rule... bs lol back to organic II have a good night man sorry I couldn't help more :(

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