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Math Help - what is x

  1. #1
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    what is x

    for what value of x is this possible
    4^x + 6^x=9^x
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  2. #2
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    Since this is posted under "Number Theory" is x assumed to be a positive integer? If so, there is no such x. 4^x+ 6^x< 9^x for all x> 1. since 4^1+ 6^2= 10> 9^1, there is a root between 1 and 2.
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  3. #3
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    Hello, jashansinghal!


    For what value of x is this possible? . 4^x + 6^x \:=\:9^x
    We have: . 9^x - 6^x - 4^x \:=\:0

    Divide by 4^x\!:\quad \frac{9^x}{4^x} - \frac{6^x}{4^x} - \frac{4^x}{4^x} \:=\:0 \quad\Rightarrow\quad \left(\frac{9}{4}\right)^x - \left(\frac{6}{4}\right)^x - 1 \;=\;0

    . . \left[\left(\frac{3}{2}\right)^2\right]^x - \left(\frac{3}{2}\right)^x - 1 \;=\;0 \quad\Rightarrow\quad \left[\left(\frac{3}{2}\right)^x\right]^2 - \left(\frac{3}{2}\right)^x - 1 \;=\;0

    . . . We have a quadratic in \left(\frac{3}{2}\right)^x


    Quadratic Formula: . \left(\frac{3}{2}\right)^x \;=\;\frac{1 \pm\sqrt{5}}{2}

    Since \left(\frac{3}{2}\right)^x > 0 , the only root is: . \left(\frac{3}{2}\right)^x \:=\:\frac{1+\sqrt{5}}{2} \:=\:\phi, the Golden Mean.


    So we have: . \left(\frac{3}{2}\right)^x \:=\:\phi \quad\Rightarrow\quad (1.5)^x \:=\:\phi


    Take logs: . \ln(1.5)^x \:=\:\ln(\phi) \quad\Rightarrow\quad x\ln(1.5) \:=\:\ln(\phi)


    Therefore: . x \;=\;\frac{\ln(\phi)}{\ln(1.5)} \;=\;1.18681439\hdots

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