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Very simple log problem - I'm missing something.

  1. Sep 7, 2014 #1
    A=12^(1/5)
    Log(A)=(1/5)Log(12)

    The next part in the book says:

    Log(A)=0.2158
    A=1.644

    I don't know how to do this without a calculator, and with a calculator I'm getting 1.24.

    I'm doing e^0.2158

    Can anyone please tell me where I'm going wrong here?


    Note: this isn't homework, just something I've come across
     
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2
    How old is the book? If old (pre 1980) then the expectation is that you have and use a set of log tables.

    Also log(A) implies log to base 10 i.e. 10.1258.
    The standard notation for logs to base e is Ln(A).
     
  4. Sep 7, 2014 #3

    jedishrfu

    Staff: Mentor

    Your mistake is using e and not 10. Base e is used when doing stuff with natural logs ie ln() and 10 is used when doing base 10 logs.

    Google can supplement a calculator if one isn't readily available.

    The old fashioned way of solving log problems was to use precomputed log tables. Alternatively you could use the ll scales of a decitrig sliderule.
     
  5. Sep 7, 2014 #4
    Thank you both! I figured the book was old because it was asking me to do this, but I still knew my log knowledge was off somewhere. I'll remember the base 10 thing in future. Thanks again!
     
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