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  1. #1
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    indices

    i have been trying to work this question but i end up with 3^{x+3} - 3 ^{x+1}

    can someone please me what i am doing wrong.
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  2. #2
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    9^{x+2} = 9^2 \cdot 9^x. Similarly, 3^{x+1} = 3 \cdot 3^x and 3^{x-1} = 3^{-1} \cdot 3^x

    Which gives \dfrac{81 \cdot 9^x - 9^x}{3 \cdot 3^x - \dfrac{1}{3} \cdot3^x} = 10(3 \cdot 3^x)

    Factor and use the fact that 9 = 3^2
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  3. #3
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    thanks for the direction, but i have tried but i seems to be doing something wrong. should i factor out the 9 from the numerator?
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    Quote Originally Posted by e^(i*pi) View Post
    9^{x+2} = 9^2 \cdot 9^x. Similarly, 3^{x+1} = 3 \cdot 3^x and 3^{x-1} = 3^{-1} \cdot 3^x

    Which gives \dfrac{81 \cdot 9^x - 9^x}{3 \cdot 3^x - \dfrac{1}{3} \cdot3^x} = 10(3 \cdot 3^x)

    Factor and use the fact that 9 = 3^2
    \dfrac{(81-1) \cdot 9^x}{3^x \left(3-\dfrac{1}{3}\right)} = \dfrac{80 \cdot 9^x}{\dfrac{8}{3} \cdot 3^x} = 30 \cdot 10^{x}

    Once you do some cancelling you get:

    \dfrac{9^x}{3^x} = 10^{x}

    You know that 9^x = (3^2)^x = 3^{2x} so rewrite the equation as

    \dfrac{3^{2x}}{3^x} = 3^x

    3^x = 10^x is trivial and by inspection only x=0 would work but you can verify using logarithms
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  5. #5
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    Hello, sigma1!

    \text{Simplify: }\:\dfrac{9^{x+2} - 9^x}{3^{x+1} - 3^{x-1}}

    \text{Numerator: }\;9^{x+2} - 9^x \:=\:(3^2)^{x+2} - (3^2)^x \:=\:3^{2x+4} - 3^{2x}

    . . . . . . . . =\;3^{2x}(3^4 - 1) \;=\;80\cdot3^{2x}


    \text{Denominator: }\;3^{x+1} - 3^{x-1} \;=\;3^{x-1}(3^2-1) \;=\;8\cdot3^{x-1}


    \text{The problem becomes: }\;\dfrac{80\cdot3^{2x}}{8\cdot3^{x-1}} \;=\;10\cdot3^{2x-(x-1)} \;=\;10\cdot3^{x+1}

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  6. #6
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    thanks alot i understood your steps very clearly...
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  7. #7
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    Quote Originally Posted by sigma1 View Post
    i have been trying to work this question but i end up with 3^{x+3} - 3 ^{x+1}

    can someone please me what i am doing wrong.
    \displaystyle\frac{9^{x+2}-9^x}{3^{x+1}-3^{x-1}}=\frac{9^x\left(9^2-1\right)}{3^{x-1}\left(3^2-1\right)}=\frac{3^{2x}\left(9^2-1\right)}{3^{x-1}\left(3^2-1\right)}

    I think you are getting your answer as follows....

    \displaystyle\frac{9^{2x}-9^x}{3^{x+1}-3^{x-1}}=\frac{3^{2x+4}-3^{2x}}{3^{x+1}-3^{x-1}}

    Now you may be subtracting x+1 from 2x+4 and x-1 from 2x which is incorrect.
    You can only do it that way if you have factors.
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