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Math Help - indices

  1. #1
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    Question indices

    Have homework on indices as follows but unable to solve. Any help is appreciated.

    Evaluate
    1) 12^x/3^x-1

    2) 12^x/3^x-1 multiply 4^x+1/2

    3) 24^x//2^3x+1 multiply [3^x+1/2^2x+1]^-2

    thank you so much
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  2. #2
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    Quote Originally Posted by kristyneo View Post
    Have homework on indices as follows but unable to solve. Any help is appreciated.

    Evaluate
    1) 12^x/3^x-1

    2) 12^x/3^x-1 multiply 4^x+1/2

    3) 24^x//2^3x+1 multiply [3^x+1/2^2x+1]^-2

    thank you so much
    Hello,

    to #1: I assume that you mean:

    \frac{12^x}{3^{x-1}}=\frac{4^x \cdot 3^x}{\frac13 \cdot 3^x} . I'm sure you can continue now

    to #2: I assume that you mean:

    \frac{12^x}{3^{x-1}} \cdot 4^{x+\frac12}= \frac{4^x \cdot 3^x \cdot 4^x \cdot 4^{\frac12}}{\frac13 \cdot 3^x} . Simplify 4^{\frac12} = \sqrt{4}=2 and then continue.

    to #3: I can't read the last problem. Please use brackets to differ between bases, exponents, summands and factors.
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  3. #3
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    Thank you for your prompt response.

    1) for 1, I cancel out 3x on top and bottom and I got 4x/which gives
    4 x .3

    Am i correct?

    2) I realise I type wrongly for 2. the question is
    (12^x)/3^(x-1) multiply (4^(x+1))/2

    3) sorry, typo error again, it should be

    24^x/2^(3x+1) multiply [3^(x+1)/2^(2x+1)]^-2
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  4. #4
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    Quote Originally Posted by kristyneo View Post
    Thank you for your prompt response.

    1) for 1, I cancel out 3x on top and bottom and I got 4x/which gives
    4 x .3

    Am i correct?

    2) I realise I type wrongly for 2. the question is
    (12^x)/3^(x-1) multiply (4^(x+1))/2

    3) sorry, typo error again, it should be

    24^x/2^(3x+1) multiply [3^(x+1)/2^(2x+1)]^-2
    Hello,

    I believe that you mean: 4^x \cdot 3 which is correct. I would write: 3 \cdot 4^x only to have the constant factor in front.

    to #2 (new version):
    \frac{12^x}{3^{x-1}} \cdot \left(4^{x+1}\right)^{\frac12}= \frac{4^x \cdot 3^x \cdot 4^{\frac12 x} \cdot 4^{\frac12}}{\frac13 \cdot 3^x} . Consider that 4^x = (2^2)^x=2^{2x} and 4^{\frac12 x}=(2^2)^{\frac12 x}=2^x . Cancel equal terms and collect like terms.

    to #3:
    \frac{24^x}{2^{3x+1}}\cdot \left(\frac{3^{x+1}}{2^{2x+1}} \right)^{-2} = \frac{3^x \cdot 8^x}{2 \cdot (2^3)^x}\cdot \left(\frac{2^{2x+1}}{3^{x+1}} \right)^{2} = \frac{3^x \cdot 8^x}{2 \cdot 8^x}\cdot \left(\frac{2^2 \cdot 2^{4x}}{3^2 \cdot 3^{2x}} \right) . To continue cancel equal terms and collect like terms.
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  5. #5
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    Hello, kristyneo!

    I have to guess what you meant . . .


    1)\;\frac{12^x}{3^{x-1}}
    We have: . \frac{(4\cdot3)^x}{3^{x-1}} \;\;=\;\frac{4^x\cdot3^x}{3^{x-1}} \;\;=\;\;4^x\cdot3 \;\;=\;\;3\!\cdot\!4^x


    2)\;\frac{12^x}{3^{x-1}} \cdot \frac{4^{x+1}}{2}
    We have: . \frac{(2^2\cdot3)^x}{3^{x-1}} \cdot \frac{(2^2)^{x+1}}{2}\;\;=\;\;\frac{2^{2x}\cdot3^x  }{3^{x-1}} \cdot\frac{2^{2x+2}}{2^1}

    . . =\;\;\frac{2^{2x}\cdot2^{2x+2}}{2^1}\cdot\frac{3^x  }{3^{x-1}} \;\;=\;\;2^{3x+1}\cdot3 \;=\;3\cdot 2^{3x+1}



    3)\;\frac{24^x}{2^{3x+1}} \cdot \left[\frac{3^{x+1}}{2^{2x+1}}\right]^{-2}

    We have: . \frac{(2^3\cdot3)^x}{2^{3x+1}}\cdot \frac{(3^{x+1})^{-2}} {(2^{2x+1})^{-2}} \;=\;\frac{2^{3x}\cdot3^x}{2^{3x+1}} \cdot\frac{3^{-2x-2}}{2^{-4x-2}} \;= \;\frac{2^{3x}\cdot 3^x\cdot3^{-2x-2}}{2^{3x+1}\cdot2^{-4x-2}}

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

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  6. #6
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    Thank you all,
    You have been a great help !!!
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