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anonymous

  • 4 years ago

solve the equation ln(X-3)=ln(X+2)+3 how do you cancel ln. I know when you have log you can do this= log base 5 (2X+4)=3

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  1. anonymous
    • 4 years ago
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    Just a start and i'm not definitive on this but can you do ln( (x-3)/(x+2) ) = 3 as subtracting logs equates to dividing them as one 'ln'.

  2. TuringTest
    • 4 years ago
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    ^that is correct then raise e to the power of both sides...

  3. anonymous
    • 4 years ago
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    can you help on this problem, Ln(x-3)=Ln(x+2)+3 please! so i can see it? how you do it

  4. TuringTest
    • 4 years ago
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    As is stated above remember the log rule\[\log a-\log b=\log(\frac ab)\]so here we can do the following:\[\ln(x-3)=\ln(x+2)+3\]\[\ln(\frac{x-3}{x+2})=3\]now raise e to the power of both sides and isolate x:\[\frac{x-3}{x+2}=e^3\]\[x-3=e^3x+2e^3\]\[x-e^3x=3+2e^3\]\[x(1-e^3)=3+2e^3\]\[x=\frac{3+2e^3}{1-e^3}\]please pay try to follow how I did this so you can do the problems yourself in the future.

  5. TuringTest
    • 4 years ago
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    please try*

  6. anonymous
    • 4 years ago
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    awesome!! thx you yes, i will try myself so i can be ready for my test, right practice take perfection.thank you so much!

  7. TuringTest
    • 4 years ago
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    welcome!

  8. anonymous
    • 4 years ago
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    LnX+Ln(X+2)=2 this is what i did Ln(X+0)/(X+2)=2 (X+0)/(X+2)=e^2 x+=e^2X+2e^2 x+e^2-x=-0+2e^2 x=2e^2/x+e^2= the calculator says error cause i can not divide by o, off course, but my professor book answer is x=1.896 or x=-3.896 what iam doing wrong?

  9. TuringTest
    • 4 years ago
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    you have a slightly different rule applying with addition\[\log a+\log b=\log(ab)\]so here you have\[\ln x+\ln(x+2)=\ln(x^2+2x)\]try to take it from there

  10. TuringTest
    • 4 years ago
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    ...and you never need to add a zero like that. No point in it.

  11. TuringTest
    • 4 years ago
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    Basic log rules:\[\log a+\log b=\log(ab)\]\[\log a-\log b=\log(\frac ab)\]\[\log(x^a)=a\log x\]\[\log_b x=\frac{\log_a x}{\log_a b}\]these are your best friends for these kind of problems.

  12. anonymous
    • 4 years ago
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    so, Ln x cancel and just end up with(x+2)/(x^2+2x)?

  13. anonymous
    • 4 years ago
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    oh.. did you multiple (x).(x+2) and end up x^2+2x-2=0 and from then do the quadratic formula?

  14. TuringTest
    • 4 years ago
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    \[\ln x+\ln(x+2)=2\]\[\ln(x^2+2x)=2\]\[x^2+2x=e^2\]\[x^2+2x-e^2=0\]now you can use the quadratic formula.

  15. anonymous
    • 4 years ago
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    thx! how about this problem e^x=35

  16. anonymous
    • 4 years ago
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    Take ln of both sides as ln undoes e^x.. so: ln(e^x) = ln(35) x = ln(35)

  17. anonymous
    • 4 years ago
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    oh i was doing e^35 iam a little confused i though the ln cancel in both sides

  18. anonymous
    • 4 years ago
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    how would you do this problem 2^3x=16 i know base is 2

  19. anonymous
    • 4 years ago
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    I'm not 100% sure but if you take log(base 2) on both sides you get 3x = 4 hence x = 4/3 is that correct?

  20. anonymous
    • 4 years ago
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    no, is 2 to the power of 3x=16

  21. anonymous
    • 4 years ago
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    2^3x=16

  22. anonymous
    • 4 years ago
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    yeah, that's what I did - solve that for x which gives you 4/3. type log base 2 (16) into your calculator and you'll get 4. hence 3x = 4

  23. anonymous
    • 4 years ago
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    ah.. thx! what the formula for that problem?

  24. anonymous
    • 4 years ago
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    None - just knowing the rules of logarithms.

  25. anonymous
    • 4 years ago
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    i started this problem original problem is 5e^4x=500 so what i did is the following:5/5 e^4x=500/5 e^4x=e^100 4x=4.60517-4 -4x x=.60517?

  26. anonymous
    • 4 years ago
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    Chances are that's incorrect as answers rarely come out like that. I think this is correct: e^4x = 500/5 e^4x = 100 ln(e^4x) = ln(100) 4x = ln(100) x = ln(25) <-- always leave your answer in terms on ln as it is more precise than a decimal.

  27. anonymous
    • 4 years ago
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    I've made a mistake again.. I'm getting tired: x = ln(100) / 4

  28. anonymous
    • 4 years ago
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    I understand you been tired, I been home work since 11a.m i shall also take a break. thank you so much for you help i really

  29. anonymous
    • 4 years ago
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    No worries.

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