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Math Help - 3 examples

  1. #1
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    3 examples

    I cannot solve these 3 examples. If you know it, please help me.
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  2. #2
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    To simplify calculations,
    for (1), begin by factorising 35 to write it as a multiple of the single-digit terms, 5 and 7, may be simpler ways though.

    For (2), 9 is a power of 3, 4 is a power of 2 and 6 is a multiple of 3 and 2.

    For (3), 4 and 8 are powers of 2, 12 is a multiple of 2, but this is a quadratic in disguise.
    Last edited by Archie Meade; January 12th 2010 at 06:35 AM.
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  3. #3
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    I understand what you said but what to do after it?
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  4. #4
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    Well, for (1)

    simplest is to write 5^x+(35)7^x=7^x+(35)5^{2x}

    If x=0, then

    5^0+(35)7^0=7^0+(35)5^0

    1+35=1+35

    However, I doubt that this is the way you are expected to solve,
    since this answer is by inspection,
    it is not a derived solution.
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  5. #5
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    5^x+(35)7^x=7^x+(35)5^{2x}

    \frac{5^x}{35}-5^{2x}=\frac{7^x}{35}-7^x

    5^{2x}(\frac{5^{-x}}{35}-1)=7^x(\frac{1}{35}-1)


    These are equal if x=0
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  6. #6
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    The third is really simple.
    Don't forget to find a domain of the function.
    2^{2(x- \sqrt{x^2 - 5})} - 12 \cdot 2^{x- \sqrt{x^2 - 5}} + 8 = 0

    2^{x- \sqrt{x^2 - 5}}=a and a>0
    a^2 - 12a + 8 = 0
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  7. #7
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    For Q2

    (6)4^{x+1}+\frac{9^{x+1}}{2}=\frac{9^{x+2}}{3}+(3)  4^x

    (6)2^{2(x+1)}+\frac{3^{2(x+1)}}{2}=3^{2(x+2)-1}+(3)2^{2x}

    (6)(4)2^{2x}+\frac{3^{2x+3}}{6}=3^{2x+3}+(3)2^{2x}

    continue
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  8. #8
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    2. Mulitply two sides of the equation throughout 6.
    After simple transformation:
    36 \cdot 4^{x+1}=2 \cdot 9^{x+2} - 3 \cdot 9^{x+1} +18 \cdot 4^x
    9 \cdot 4^{x+2} = 2 \cdot 9^{x+2} - 3 \cdot 9^{x+1} + 3 \cdot 4^x
    4^x(4^2 \cdot 9  - 18) = 9^x(2 \cdot 9^2 - 3 \cdot 9)
    (\frac{2}{3})^x = \frac{135}{126}

    If I didn't make a mistake to give a result you need to use a logarithm.
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  9. #9
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    Hello, x-mather!

    These are truly ugly problems . . .


    2)\;\;6\cdot4^{x+1} \;=\;\frac{9^{x+2}}{3} - \frac{9^{x+1}}{2} + 3\cdot4^x

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

    . . . . . . . . . 2\cdot3\cdot2^{2x+2} \;=\;\frac{3^{2x+4}}{3} - \frac{3^{2x+2}}{2} + 3\cdot2^{2x}

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

    Multiply by 2: . 3\cdot2^{2x+4} \;=\;2\cdot3^{2x+3} - 3^{2x+2} + 3\cdot2^{2x+1}

    . . . . 3\cdot2^{2x+4} - 3\cdot2^{2x+1} \;=\;2\cdot3^{2x+3} - 3^{2x+2}


    Factor: . 3\cdot2^{2x+1}(2^3-1) \;=\;3^{2x+2}(2\cdot3-1)

    . . . . . . . . . 3\cdot2^{2x+1}\cdot7 \;=\;3^{2x+1}\cdot5


    Divide by 3: . . 7\cdot2^{2x+1} \;=\;5\cdot3^{2x}

    . . . . . . . . . . . . \frac{2^{2x+1}}{3^{2x}} \;=\;\frac{5}{7}

    Divide by 2: . . . . \frac{2^{2x}}{3^{2x}} \;=\;\frac{5}{14}

    . . . . . . . . . . . \left(\frac{2}{3}\right)^{2x} \;=\;\frac{5}{14}


    Take logs: . . \ln\left(\tfrac{2}{3}\right)^{2x} \;=\;\ln\left(\tfrac{5}{14}\right)

    . . . . . . . . . 2x\ln\left(\tfrac{2}{3}\right) \;=\;\ln\left(\tfrac{5}{14}\right)


    . . Therefore: . . . . x \;=\;\frac{\ln\left(\frac{5}{14}\right)}{2\ln\left  (\frac{2}{3}\right)}

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  10. #10
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    Late, while I was writing I didn't see a new post appeared.
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