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Thread: 81^x-2

  1. #1
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    Talking 81^x-2

    Question is in topic.
    You must express this in form 9^y
    I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question

    EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

    Thanks!
    Last edited by Danielisew; Oct 19th 2007 at 09:55 AM. Reason: I forgot to add another question
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  2. #2
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    Quote Originally Posted by Danielisew View Post
    Question is in topic.
    You must express this in form 9^y
    I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question

    EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

    Thanks!
    "Question is in topic." What topic?

    Also: you need to use parenthesis.

    For the first problem:
    $\displaystyle \frac{1}{9^{1 - x}} = 9^{-(1 - x)} = 9^{x - 1}$

    For the second problem:
    $\displaystyle 3^{4x+6}$

    Note that $\displaystyle 9 = 3^{1/2}$, so
    $\displaystyle 3^{4x+6} = (9^{1/2})^{4x + 6} = 9^{\frac{4x + 6}{2}} = 9^{2x + 3}$

    and what does any of this have to do with $\displaystyle 81 - x^2$???

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Danielisew View Post
    Question is in topic.
    You must express this in form 9^y
    I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question

    EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

    Thanks!
    Try putting in the missing brackets to make your meaning clear.

    RonL
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  4. #4
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    For the first one notice that $\displaystyle 81=9^{2}$

    So we get $\displaystyle 9^{2(x-2)}$

    Distribute and get $\displaystyle 9^{2x-4}$
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