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Question about logarithm

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    6^log x=1/36

    2. Relevant equations
    Y=logcx

    3. The attempt at a solution
    How do u solve this? I know the 1/36 is the exponent. Usually the logs that u normally do is not in an exponent like this. The answer is 0.01. How did they get that?
     
  2. jcsd
  3. Oct 30, 2013 #2

    Dick

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    Take the log of both sides. Use rules of logs and solve for log(x). If you want to take a shortcut, 1/36=6^(-2), right?
     
  4. Oct 30, 2013 #3
    I thought the ans was -2 since 6^-2 = 1/36

    But the ans is 0.01. So that means it is not...
     
  5. Oct 30, 2013 #4

    Dick

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    log(x)=(-2). x isn't -2. You are supposed to solve for x. What's the base of your logarithms?
     
  6. Oct 30, 2013 #5
    Is the base 10?
     
  7. Oct 30, 2013 #6

    Dick

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    It should be stated in the problems or your book what the base is. Suppose it is 10. Then if log(x)=(-2), what is x?
     
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