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Egirl01

  • 2 years ago

2log(x-3)+1 = 5 Could you break it down step by step? I'm a bit confused on how to do this algebraically

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  1. Callisto
    • 2 years ago
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    To solve x?

  2. Egirl01
    • 2 years ago
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    yes, i presume

  3. Callisto
    • 2 years ago
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    Okay, first, subtract both sides by 1, what do you get?

  4. Egirl01
    • 2 years ago
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    you get 2log(x-3) = 4

  5. Callisto
    • 2 years ago
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    Right, now divide both sides by 2, what do you get?

  6. Egirl01
    • 2 years ago
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    log(x-3) = 2

  7. Egirl01
    • 2 years ago
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    haha, that's fine :)

  8. calculusfunctions
    • 2 years ago
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    Don't mind me. @Callisto seems to be doing a fine job teaching so I'll just observe if you don't mind? I'll only respond if you ask.

  9. Callisto
    • 2 years ago
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    Yup! Now, take anti-log for both sides. What do you get? Note: anti-log = inverse of log For example, logx = 1 => 10^(logx) = 10^1 => x = 10^1 = 10

  10. Egirl01
    • 2 years ago
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    yikes, haha, i'm not for sure. I briefly remember anti-logs but I'm not for sure what do with them

  11. Callisto
    • 2 years ago
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    Do you understand that example?

  12. Egirl01
    • 2 years ago
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    i'm starting to get it. logx = 1. I understand that but not after that

  13. calculusfunctions
    • 2 years ago
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    If I may, @Callisto perhaps he'd @Egirl01 would understand if you asked her to change the logarithm form to exponential form. @Egirl01 do you know how to do that.

  14. Egirl01
    • 2 years ago
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    No, I'm sorry

  15. Egirl01
    • 2 years ago
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    give me one second

  16. ByteMe
    • 2 years ago
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    \(\large b^x=y \rightarrow log_by=x \)

  17. Egirl01
    • 2 years ago
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    log^2 = x-3?

  18. Callisto
    • 2 years ago
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    logx = 1 is actually just an example... ---------------------------------- Intuitively, log a number is the value of the power of the base Let say, I have a number 100, which can be rewritten as 10^2 Now, I take log (base 10) of that number (100), so I'll get 2, since 10 to the power of 2 give me 100. (Sorry if I make it even worse...)

  19. Egirl01
    • 2 years ago
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    No, I understand the piece. so if its simply log, then its naturally log10, and the anti-log of log10 is 1/10 right?

  20. Callisto
    • 2 years ago
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    anti-log of log 10 is 10. \[\log^{-1}(log10) = 10\] anti-log is the inverse of log.

  21. Egirl01
    • 2 years ago
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    Alright. So then it'd be 10 (x-3) = 2?

  22. Callisto
    • 2 years ago
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    No.... How did you get that?

  23. Egirl01
    • 2 years ago
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    the anti-log of log 10 is 10, so log(x-3) = 2 would turn to 10(x-3) = 2? Obviously that is wrong. I'm a bit lost but you can continue onto the next part of solving the equation. I'll eventually figure what I'm doing wrong

  24. Callisto
    • 2 years ago
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    \[log(x-3) = 2\]Take anti-log for both sides: \[\log^{-1}(log(x-3)) = \log^{-1}(2)\] Perhaps you can simplify the left first.

  25. Egirl01
    • 2 years ago
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    I still don't quite follow, but continue on. I'll understand eventually :)

  26. Callisto
    • 2 years ago
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    Nope.. It's your turn to work on it... Which part you don't understand ?

  27. Egirl01
    • 2 years ago
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    how exactly I create the anti-log from log(x-3)

  28. Callisto
    • 2 years ago
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    Because you are doing something on both sides... You create anti-log to undo the work of log...

  29. Egirl01
    • 2 years ago
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    So the anti-log of log10 is 10, right? and the anti-log of 2 is what?

  30. Callisto
    • 2 years ago
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    Before answering you question, could you please answer few questions first?

  31. Egirl01
    • 2 years ago
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    okay

  32. Callisto
    • 2 years ago
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    What is log10?

  33. Egirl01
    • 2 years ago
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    log10 =1

  34. Callisto
    • 2 years ago
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    What is log100?

  35. Egirl01
    • 2 years ago
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    log100 = 2

  36. Callisto
    • 2 years ago
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    Can we tell me what number do I have to take log on in order to get 1?

  37. Egirl01
    • 2 years ago
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    10?

  38. Callisto
    • 2 years ago
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    Yes. So, what number do I have to take log on in order to get 2?

  39. Egirl01
    • 2 years ago
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    100

  40. Callisto
    • 2 years ago
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    Yes! How did you get this answer?

  41. Egirl01
    • 2 years ago
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    because you inversed it?

  42. Callisto
    • 2 years ago
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    Yes! So, now, can you answer your question? what is \(\log^{-1}2\)?

  43. Egirl01
    • 2 years ago
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    It is 100

  44. Callisto
    • 2 years ago
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    Yes!

  45. Egirl01
    • 2 years ago
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    So 10(x-3) = 100?

  46. Callisto
    • 2 years ago
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    No. Another set question questions... What is log10 (again)?

  47. Egirl01
    • 2 years ago
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    log 10 is 1

  48. Callisto
    • 2 years ago
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    You got it right for the right side, it's wrong for the left.

  49. Callisto
    • 2 years ago
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    What is \(\log^{-1}log(10)\)?

  50. Egirl01
    • 2 years ago
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    10

  51. Callisto
    • 2 years ago
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    Yes. What is \(\log^{-1}(log100)\)?

  52. Egirl01
    • 2 years ago
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    2?

  53. Egirl01
    • 2 years ago
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    WAIT

  54. Egirl01
    • 2 years ago
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    100?

  55. Callisto
    • 2 years ago
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    Yes!

  56. Callisto
    • 2 years ago
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    What is \(\log^{-1} (log1000)\)?

  57. Egirl01
    • 2 years ago
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    1000

  58. Callisto
    • 2 years ago
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    So, we can see that \(\log^{-1}\) of log a number is the number itself, agree?

  59. Egirl01
    • 2 years ago
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    yes

  60. Callisto
    • 2 years ago
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    So, \(\log^{-1} (\log y) = y\), agree?

  61. Egirl01
    • 2 years ago
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    yes

  62. Callisto
    • 2 years ago
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    So, what is \(\log^{-1} (log(x-3))\)?

  63. Egirl01
    • 2 years ago
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    x-3?

  64. Callisto
    • 2 years ago
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    Yes!

  65. Callisto
    • 2 years ago
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    Can you simplify the following now? \[\log^{-1}(log(x-3)) = \log^{-1}(2)\]

  66. Egirl01
    • 2 years ago
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    x-3 = 100?

  67. Callisto
    • 2 years ago
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    Yes! Now can you solve it?

  68. Egirl01
    • 2 years ago
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    yes

  69. Egirl01
    • 2 years ago
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    x = 103?

  70. Callisto
    • 2 years ago
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    Nice :D and Yes! Do you understand how to solve this type of question now?

  71. Egirl01
    • 2 years ago
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    yes! Thank you soooo much for your help! I really appreciate it!

  72. Callisto
    • 2 years ago
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    You're welcome :)

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