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Simplify the expression:
The first 7 is regular size and the log7x are small (the second 7 is subscripted only).
7log7x =

would the answer be 1?

How do you work this?

  • Math+Log -

    is it


    if so, then the answer is x

    basic log rule: a^(log k, with base a) = k

  • Math+Log -

    Ahh yea it looks like that. Hey can you teach me how to do all those subscripts and those impossible math symbols?

  • Math+Log -

    27log27x = x


  • Math+Log -

    correct, if they have the same appearance as your first question.

    (As tutors some have been given special privileges that students don't have.
    for example I can post a link which normally does not work here)

  • Math+Log -


    The first 3 is a subscript

    Would that = x?

    Also with

    log15(15^x) = x ?

    log221(221^x)= x?

  • Math+Log -

    loga(a^x) = x
    for any old a
    loga (a^x) = x loga(a)
    but loga(a) = 1

  • Math+Log -

    remember that log (a^n) = n log a with any base
    log3 3^x = xlog3 3
    = x

    yes you are right.

  • Math+Log -

    ohh yea ok thanks.

    so the subscript is like the answer to 15^x? and the only way to make it = 15 is if x = 1


  • Math+Log -

    Wait so is it x or 1?

  • Math+Log -

    it is x

    look at Damon's last two lines:

    " loga (a^x) = x loga(a)
    but loga(a) = 1 "

    so wouldn't loga (a^x) = x(1) = x ?

  • Math+Log -

    oh yea it is x
    I thought it was 15x at first, but then i realize that it was its suppose to be 15^x


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