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Math Help - How to solve this powers problem

  1. #1
    Senior Member Educated's Avatar
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    How to solve this powers problem

    This equation was posted up by OnlyMath but got deleted because he posted 3 problems in 1 thread. He has not reposted it so I am because I'm curious as to how to solve it.

    \displaystyle{2(4^x) + 6^x = 9^x} Find x.

    I already know the answer is:

    \displaystyle{x = \dfrac{log2}{log(\frac{3}{2})}}

    \displaystyle{x \approx{1.70951}


    I have tried really hard to try and solve it, but I get stuck because there is a plus between the powers.
    I can't go \sqrt[x]{-6^x}+\sqrt[x]{9^x} because it doesn't work. You have to go \sqrt[x]{-6^x + 9^x} which doesn't help in solving x.

    Can someone show me a step in the right direction?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Educated View Post
    This equation was posted up by OnlyMath but got deleted because he posted 3 problems in 1 thread. He has not reposted it so I am because I'm curious as to how to solve it.

    \displaystyle{2(4^x) + 6^x = 9^x} Find x.

    I already know the answer is:

    \displaystyle{x = \dfrac{log2}{log(\frac{3}{2})}}

    \displaystyle{x \approx{1.70951}


    I have tried really hard to try and solve it, but I get stuck because there is a plus between the powers.
    I can't go \sqrt[x]{-6^x}+\sqrt[x]{9^x} because it doesn't work. You have to go \sqrt[x]{-6^x + 9^x} which doesn't help in solving x.

    Can someone show me a step in the right direction?
    Arrange the equation to get

    <br />
(3^x)^2-3^x\cdot 2^x-2(2^x)^2=0

    <br />
(3^x-2\cdot 2^x)(3^x+2^x)=0

    The second factor is undefined since a^x<0.

    As for the first factor,

    3^x=2^{x+1}

    \frac{x}{x+1}=\frac{\log 2}{\log 3}

    x=\frac{\log 2}{\log \frac{3}{2}}
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  3. #3
    Senior Member Educated's Avatar
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    Oh I got it, I would have never thought of trying to factorise it.

    But the final step seems confusing.
    Instead I did this for the final step:

    3^x=2^{x+1}

    (x)log{3}=(x+1)log{2}

    (x)log{3}=(x)log{2} + (1)log{2}

    (x)log{3}-(x)log{2}= log{2}

    x(log{3}-log{2})= log{2}

    x= \dfrac{log{2}}{(log{3}-log{2})}

    x= \dfrac{log{2}}{log{(\frac{3}{2}})}

    Thanks!
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