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Thread: Exponential equation

  1. #1
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    Exponential equation

    How can I solve for r in the equation 2x = x^r for 0<x<1?

    I did the following:
    log 2 + log x = r log x
    r = log 2 / log x + 1

    However, for the value of x = .5, r evaluates to 0, which gives you the equation 2x = 1.
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  2. #2
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    Quote Originally Posted by adnaps1 View Post
    How can I solve for r in the equation 2x = x^r for 0<x<1?

    I did the following:
    log 2 + log x = r log x
    r = log 2 / log x + 1

    However, for the value of x = .5, r evaluates to 0, which gives you the equation 2x = 1.
    what is wrong with $\displaystyle r = 0$ ? $\displaystyle x$ is the variable that has the restriction.

    $\displaystyle 2x - x^r = 0$

    $\displaystyle x(2 - x^{r-1}) = 0$

    $\displaystyle x = 0$ ... invalid solution since $\displaystyle 0 < x < 1$

    $\displaystyle x^{r-1} = 2$

    $\displaystyle \displaystyle x = 2^{\frac{1}{r-1}}$

    since $\displaystyle 0 < x < 1$ ...

    $\displaystyle \displaystyle \frac{1}{r-1} < 0$

    $\displaystyle r < 1$
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  3. #3
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    Thanks a lot, skeeter.
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