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anonymous

  • 5 years ago

i need help solving logarithmic equations

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  1. anonymous
    • 5 years ago
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    \[\log_{5} 4-\log_{5} 3r=3\log_{5} 2\]

  2. anonymous
    • 5 years ago
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    you want to find r? ^_^

  3. anonymous
    • 5 years ago
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    yes.

  4. anonymous
    • 5 years ago
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    but i need step by step help so i can understand it.

  5. anonymous
    • 5 years ago
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    Alright, here we go :) :\[-\log_{5}3r = \log_{5}2^3 - \log_{5}4 \]\[\log_{5}3r = -\log_{5}2^3 +\log_{5}4 \]\[\log_{5}3r = \log_{5}(8.4) \]\[\log_{5}3r = \log_{5}32 \]\[\log_{5}3r = 2.15 \rightarrow 3r = 5^{2.15} \rightarrow r = \frac{5^{2.15}}{3} \approx10.6 \] correct me if I'm wrong please ^_^

  6. anonymous
    • 5 years ago
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    i'm not sure if it's right or wrong. that's my problem. i have no knowledge on this at all.

  7. anonymous
    • 5 years ago
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    hmm , alright the basic rules : \[\log_{b}n + \log_{b}k = \log_{b}(n + k) \] \[ \log_{b}n -\log_{b}k = \log_{b}(\frac{n}{k})\] I think I've done a silly mistake lol , hold on :)

  8. anonymous
    • 5 years ago
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    Use the properties of logerithms: \[\log _{5}4-\log _{5}3r=\log _{5}[4/(3r)]\] Also \[3\log _{5}2=3\log _{5}2^3=3\log _{5}8\] \[\log _{5} [4/(3r)]=\log _{5}8\] 4/3r=8 So r=1/6

  9. anonymous
    • 5 years ago
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    =// after the Also ignore the 3s after the equal signs

  10. anonymous
    • 5 years ago
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    this step :\[\log_{5} 3r = \log_{5}4 - \log_{5} 8\] must be : \[\log_{5}3r = \log_{5} \frac{4}{8}\]

  11. anonymous
    • 5 years ago
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    you can take it up from here, right? :) I think xav's answer is correct since I did a silly mistake lol, sorry ^_^"

  12. anonymous
    • 5 years ago
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    r=1/6

  13. anonymous
    • 5 years ago
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    Thats correct

  14. anonymous
    • 5 years ago
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    Thank you so much :)

  15. anonymous
    • 5 years ago
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    alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?

  16. anonymous
    • 5 years ago
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    alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?

  17. anonymous
    • 5 years ago
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    Well, you can verify if the answer is correct or not by going here http://www.wolframalpha.com/input/?i=log5%284%29%E2%88%92log5%283r%29%3D3log5%282%29

  18. anonymous
    • 5 years ago
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    Yes i did. I have plenty more problems i need help with. Can anyone be patient with me and help me through all of them ? :)

  19. anonymous
    • 5 years ago
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    what are your problems?

  20. anonymous
    • 5 years ago
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    What are your problems?

  21. anonymous
    • 5 years ago
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    \[\log_{2} 4-\log_{2} (x+3)-\log_{2} 8\]

  22. anonymous
    • 5 years ago
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    What do you want to do with this

  23. anonymous
    • 5 years ago
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    Simplify ?

  24. anonymous
    • 5 years ago
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    I guess solve for x

  25. anonymous
    • 5 years ago
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    Wait... >_<

  26. anonymous
    • 5 years ago
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    Where is the =

  27. anonymous
    • 5 years ago
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    No. all the problems i'm posting are solving each equation.

  28. anonymous
    • 5 years ago
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    Where is the =

  29. anonymous
    • 5 years ago
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    oh wait i see what you saying the = is supposed to be after (x+3)

  30. anonymous
    • 5 years ago
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    okay

  31. anonymous
    • 5 years ago
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    \[2-log2(x+3)=3\]

  32. anonymous
    • 5 years ago
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    Where 2 is at the base

  33. anonymous
    • 5 years ago
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    yes 2 is the base.

  34. anonymous
    • 5 years ago
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    But that leads to no solution

  35. anonymous
    • 5 years ago
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    So instead if we use

  36. anonymous
    • 5 years ago
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    \[\log_{2} 4-\log_{2} 8=\log_{2} (x+3)\]

  37. anonymous
    • 5 years ago
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    We can write (4/8)=x+3

  38. anonymous
    • 5 years ago
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    The rest you can do

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