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Exponential Equation Problem

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data

    (2^(x+1)) / (5^x)) = 3


    2.The attempt at a solution

    (2^(x+1)) / (5^x)) = 3

    ((log 2 (x+1)) / (x log 5)) = log 3

    (x log 2 + log 2) / (x log 5) = log 3

    ( log (2^ (x+1))) / (log (5^x)) = log 3

    log (base: 5^x) (number: 2 ^ (x+1)) = log 3

    (5^x) ^ (log 3) = 2 ^ (x+1)

    (x log 5) ( log 3) = x log 2 + log 2

    x ~ 9.09

    I know that this is not the right answer but I don't understand where I went wrong.
     
  2. jcsd
  3. Jan 5, 2014 #2

    tiny-tim

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    Hi ThatDude! :smile:

    (try using the X2 button just above the Reply box :wink:)
    ((log 2 (x+1)) minus (x log 5)) = log 3 :smile:

    btw, it would have been simpler to start by expressing the LHS as a multiple of (2/5)x :wink:
     
  4. Jan 5, 2014 #3
    Thank you for helping me out.

    I thought that you could only subtract two logs when the number inside the log was being divided. In this case, isn't it the division of two logs, albeit with the same base?

    Ex:
    = log5 x/12

    = log5x - log512

    But from what I understand from your post, the following is also true:


    log5 x/12 = log5x / log5 12
     
  5. Jan 5, 2014 #4

    tiny-tim

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    it is

    you confused yourself by leaving out a step …

    2x+1 / 5x = 3

    log (2x+1 / 5x) = log 3

    log (2x+1) - log (5x) = log 3 :wink:
     
  6. Jan 5, 2014 #5

    SteamKing

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    You have some confusing expressions:

    What does log5 x/12 mean?

    Is it log (5x/12) or log[itex]_{5}(x/12)[/itex]?
     
  7. Jan 5, 2014 #6

    Ray Vickson

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    It is impossible to say whether you are correct, or not, because of the poor notation. If ##\log 5^{x/12}## means ##\log_5 (x/12)## then what you wrote is true. If it means ##\log \left( 5^{x/12}\right)## then what you wrote is false.
     
  8. Jan 5, 2014 #7
    Ah. I see. Thank you for your help.

    Yes... Ray Vickson and SteamKing, it is log5(x/12) ; that is, the base is 5 and the argument is x divided by 12. I apologize for the poor notation.
     
  9. Jan 5, 2014 #8

    Ray Vickson

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    In that case you should write something like log[5] x/12 or log_5 x/12 or log5 x/12; just about everybody would "get it" if you wrote it in any of these three ways (although it would not hurt to also include a brief verbal description, saying what the '5' means). It would be even better to include parentheses, like this: log[5] (x/12) or log_5 (x/12) or log5 (x/12).
     
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