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Math Help - Log question

  1. #1
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    Log question

    Hey guys.. I am doing some correspondence work and I was wondering if I could have help with a question. Am I just supposed to do this by trial and error? Seems like there should be an easier way.

    http://s180.photobucket.com/albums/x...=logquest2.jpg

    Thanks
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  2. #2
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    Logs

    Hello slipery

    The definition of a logarithm of a certain number is that power to which the base must be raised to give you the number. So:

    If x = \log_ay, then a^x = y.

    Can you see it now?

    Grandad
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  3. #3
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    Hi Slipery,

    You can calculate that value using the following;

    y = log7(12)

    7^y = 12

    y*ln(7) = ln(12)

    y = ln(12)/ln(7)

    Therefore, log7(12) = ln(12)/ln(7)

    This is also known more generally as the 'Change of base formula'.

    logA(B) = ln(B)/ln(A)

    EDIT: I apologize, I misread the picture you linked - feel free to ignore me.
    Last edited by Revenantus; January 8th 2009 at 09:23 AM. Reason: Misread question user was asking.
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  4. #4
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    I don't think you misread my post Revenantus.
    And I do understand what you're saying Grandad.

    The method you laid out Revantus is the one that I would do myself. I am just wondering whether that is correct.

    It is basically saying 7^ log7(12).

    First I must figure out log 7(12), and then all I do is work that out as an exponenet for 7. Log 7(12) is what I am having trouble with. I am aware of how to use the scientific calculator to figure the answer out, but none of the examples I am given seem to come out in a decimal like that, so I am just wondering if that is indeed how I am supposed to solve the problem
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  5. #5
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    Instead of just the mathematical notation, let's think through what each part means.

    log7(12) means "The power I must raise 7 to in order to result in 12".

    Therefore 7^log7(12) means "I raise 7 to the power that I must raise 7 to in order to result in 12"
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  6. #6
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    So it's obviously going to be a long decimal number
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  7. #7
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    Not quite. I'll try another example.

    e.g. 2^log2(4)

    This expression says that I'm raising 2 to the power that it needs to be raised to in order to result in 4.

    So, 2^log2(4) = 4

    We can verify this easily because log2(4)=2 and 2^2=4.
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  8. #8
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    Yes, but that works out easily because your using 2 and 4... Using 7 and 12 will still result in a decimal, will it not?
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  9. #9
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    Another approach would be to look back at the two definitions Grandad gave us.

    a^x = y and x=loga(y)

    By substituting loga(y) in place of x in the first definition we can form;

    a^loga(y) = y

    Your expression is of the same form as above, try plugging in the values and see what you get.
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  10. #10
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    Ahhh, so I worked it out and got 12 as an answer, similar to that equation..
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  11. #11
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    I understand now.. Its 7 to the 'whatever exponenet it takes 7 to get to 12'.. so obviously my original answer will be 12 as well..

    Thanks a lot Revenantus and Grandad
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  12. #12
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    Excellent, well done.
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  13. #13
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    Logs

    Right!

    Grandad
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