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Math Help - Solving Exponential Equations

  1. #1
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    Question Solving Exponential Equations

    <br />
\frac{2^x + 2^(-x)}{2}=3<br />
    I honestly have no clue how to do this because I was absent when we went over it. Please help!
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  2. #2
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    Hello,
    Quote Originally Posted by LoveBeachForever View Post
    <br />
\frac{2^x + 2^(-x)}{2}=3<br />
    I honestly have no clue how to do this because I was absent when we went over it. Please help!
    Let y=2^x

    and remember that a^{-b}=\frac{1}{a^b}
    hence 2^{-x}=\frac{1}{2^x}
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  3. #3
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    Hello, LoveBeachForever!

    Solve: . \frac{2^x + 2^{-x}}{2}\:=\:3

    Multiply by 2: . 2^x + 2^{-x} \;=\;6

    Multiply by 2^x\!:\;\;2^{2x} + 1 \;=\;6\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 6\!\cdot\!2^x + 1 \:=\:0

    So we have: . (2^x)^2 - 6(2^x) + 1 \;=\;0

    Let u \,=\,2^x and we have: . u^2 - 6u + 1 \:=\:0

    The Quadratic Formula gives us: . u \;=\;3 \pm2\sqrt{2}

    Back-substitute: . 2^x \;=\;3\pm2\sqrt{2}

    Take logs: . \ln(2^x) \;=\;\ln(3 \pm2\sqrt{2}) \quad\Rightarrow\quad x\ln(2) \;=\;\ln(3 \pm2\sqrt{2})


    Therefore: . x \;=\;\frac{\ln(3 \pm2\sqrt{2})}{\ln(2)} \;\approx\; \pm 2.543106606

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  4. #4
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    Thanks!!

    I thank both of you so much, I understand this now!
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