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Math Help - 9^x = 81. Determine x.

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    9^x = 81. Determine x.

    9^x = 81. Determine x.

    Now I found that the answer is 2 via trial and error. However, I wanted to know whether or not there was a formal mathematical procedure which proves my answer. I haven't been taught this unit yet, for the unit I'm working on now (involving inverse functions) hasn't been assigned yet. I've just been working ahead.
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    Quote Originally Posted by shadow6 View Post
    9^x = 81. Determine x.

    Now I found that the answer is 2 via trial and error. However, I wanted to know whether or not there was a formal mathematical procedure which proves my answer. I haven't been taught this unit yet, for the unit I'm working on now (involving inverse functions) hasn't been assigned yet. I've just been working ahead.
    1. 9^x = 81 = 9^2

    If the bases are equal the exponents must be equal so x = 2

    2. x\, \log_9(9) = \log_9 (81). Hence x=2
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    "Trial and Error" is a perfecly good "mathematical procedure". What more do you want?
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    Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

    8^m = 256, determine m.

    The answer is m = \frac {8}{3}.

    How would I determine the answer of this question, without using the back of the book?
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by shadow6 View Post
    Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

    8^m = 256, determine m.

    The answer is m = \frac {8}{3}.

    How would I determine the answer of this question, without using the back of the book?
    8^m = 256

    {(2^3)^m = 2^8}

    2^{3m} = 2^8

    therefore, by the equality of bases, you get

    3m = 8

    m = \frac{8}{3}

    Is that clear enough?
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by shadow6 View Post
    Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

    8^m = 256, determine m.

    The answer is m = \frac {8}{3}.

    How would I determine the answer of this question, without using the back of the book?
    This probably isn't the most general method, but

    8^m=256=2^8

    (2^3)^m=2^8

    2^{3m}=2^8

    3m=8

    m=\frac{3}{8}

    Since I do a bit of computer programming, some powers of 2 are very easy to recognize... someone will probably post a more general method after me, but tricks like this can often be used (express the base, in this case 8, as a smaller base raised to a power, in this case 2^3).
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    Yes, that's exactly what I'm looking for! Thx!
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  8. #8
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    If the problem does not come out so evenly, like maybe 3^x=17, you can just take the log (to any base) of both sides and solve for x. So, if a and b are constants, and

    a^x=b

    the solution is:

    \log{a^x}=\log{b}

    x\log{a}=\log{b}

    x=\frac{\log{b}}{\log{a}}

    Of course, that's more work than you need to do for a case like 9^x=81.

    - Hollywood
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