College Algebra/pre-calculus

The question is to evaluate each logarithm:

My professor wants us to show work. Could you please tell me if I'm on the right track? Thank you!!!

1. log6^(1/36)= -2

2. ln(1/e^4)= 1/e^2

3. log7^1= 0

4. log4^64= ln64/ln4= 3

5. log10^8= ln8/ln10= .9031

6. log3^(radical 3)= I wasn't sure how to show the work on this one. I plugged it into my calculator and got .8263

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  1. #1
    How did you get -2 ??

    log 6^(1/36)
    = (1/36) log 6 = appr .0216

    = ln1 - ln e^4
    = 0 - 4lne
    = 0 - 4(1) = -4

    log 7^1
    = log 7 = appr .845

    log 4^64
    = 64 log4 = appr 38.532

    log 10^8
    = 8log10
    = 8

    #6 --- the only one you got right

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    posted by Reiny
  2. Some of these blokes use

    log6^(1/36) to mean log6(1/36), which is indeed -2

    #2 ln(1/e^4) = ln(e^-4) = -4
    #3 ok
    #4 ok
    #5 since the other logs specify a base, this must be just a common log, and log 10^8 = 8
    #6 log3√3 = log33^(1/2) = 1/2

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    posted by Steve
  3. Hi thank you so much for helping me. Yeah, I am sorry. I do not know how to make the base numbers small like that. Thank you both again for the help!

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  4. Thanks Steve
    Had an inkling that's what he was doing, but then it was contradicted in how he worked the last one.
    btw, was it overkill in my solution to

    Almost gave up on it, but found some clues on a webpage.

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    posted by Reiny
  5. The subscripts are a trick of HTML. Most of the codes do not work on this site, but the sub and /sub tags work.

    For superscripts, the sup and /sup tags work.

    enclose the words in <> brackets to make them into HTML tags.

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    posted by Steve

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