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Math Help - Log question. (without a calc evaluate)

  1. #1
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    Log question. (without a calc evaluate)

    Hi there,

    I have a question, how do you evaluate the following without a calculator?

    log2(log3(log5^125)))

    Thanks

    ps. log2, log3 and log5 are the bases
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  2. #2
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    Quote Originally Posted by captain_jonno View Post
    Hi there,

    I have a question, how do you evaluate the following without a calculator?

    log2(log3(log5^125)))

    Thanks

    ps. log2, log3 and log5 are the bases
    Err doesn't really make sense.

    Do you mean:

     \log_2(\log_3(\log_5(125))) ?

    If so, take the very first term, and set it equal to y:

     y = \log_5(125)

    This means that  5^y = 125

    You should be able to tell that  125 = 5^3 , and hence  y = 3 .

    So now you have:

     \log_2(\log_3(3))

    Now take that term again, and set it equal to another variable, x.

     x = \log_3(3)

     3^x = 3 . Clearly x = 1

    This gives:

     \log_2(1)

    Set it equal to a random variable:

     z = \log_2(1)

     2^z = 1

    Clearly  z = 0
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  3. #3
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    Quote Originally Posted by Mush View Post
    Err doesn't really make sense.

    Do you mean:

     \log_2(\log_3(\log_5(125))) ?
    Indeed I do, though there is no front bracket '(' on there 125 for my question.
    it says: Without a calculator, evaluate the following:
     \log_2(\log_3(\log_5125)))
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  4. #4
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    Quote Originally Posted by captain_jonno View Post
    Indeed I do, though there is no front bracket '(' on there 125 for my question.
    See my edited post above.
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  5. #5
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    Thanks a bunch Mush, very quick response.

    so its just 0(1(3)))
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  6. #6
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    Quote Originally Posted by captain_jonno View Post
    Thanks a bunch Mush, very quick response.

    so its just 0(1(3)))
    Nah it's more like:

     \log_2(\log_3(\log_5125)))

     = \log_2(\log_3(3))

     = \log_2(1)

     = 0
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  7. #7
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    Thanks, thats actually how i wrote it down in the end . good stuff
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