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Math Help - solve in logs to base 10

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    solve in logs to base 10

    (2^x) - 6/(2^-x)=6
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    Re: solve in logs to base 10

    Quote Originally Posted by sluggerbroth View Post
    (2^x) - 6/(2^-x)=6
    \displaystyle \begin{align*} 2^x - \frac{6}{2^{-x}} &= 6 \\ 2^x - 6\cdot 2^x &= 6 \\ -5\cdot 2^x &= 6 \\ 2^x &= -\frac{6}{5} \end{align*}

    This does not have any solutions in the real numbers, because \displaystyle \begin{align*} 2^x > 0 \end{align*} for all \displaystyle \begin{align*} x \end{align*} .
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    Re: solve in logs to base 10

    Quote Originally Posted by sluggerbroth View Post
    (2^x) - 6/(2^-x)=6
    If the problem is 2^x-\frac{6}{2^{-x}}=6 then there is no solution.

    If that is not the question, then what is it?
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    Re: solve in logs to base 10

    Hello, sluggerbroth!

    (2^x) - 6/(2^-x) = 6

    I suspect that the problem is: . 2^x - \frac{6}{2^x} \:=\:6

    Multiply by 2^x:\;(2^x)^2 - 6 \:=\:6\cdot2^x

    And we have: . (2^x)^2 - 6\cdot2^x - 6 \:=\:0 . . . a quadratic in 2^x

    Quadratic Formula: . 2^x \:=\:\frac{6\pm\sqrt{60}}{2} \:=\:3\pm\sqrt{15}

    Take logs, base 10: . \log(2^x) \:=\:\log(3\pm\sqrt{15})

    . . . . . . . . . . . . . . x\cdot\log(2) \:=\:\log(3\pm\sqrt{15})

    n . . . . . . . . . . . . . . . . . . \boxed{\;x \:=\:\frac{\log(3 + \sqrt{15})}{\log(2)}\;}

    Note that 3 - \sqrt{15} is negative.
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