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Thread: Exponential equation help

  1. #1
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    Exponential equation help

    Need help with problem. I think it is a characteristic equation where u let y= something

    5 2x-1 + 5x=10
    Last edited by Nfalconer; Jul 7th 2018 at 09:25 PM.
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  2. #2
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    Re: Exponential equation help

    Is the equation:

    $\displaystyle 5^{2x-1}+5^x=10$ ?
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    Re: Exponential equation help

    Yes that is the equation. Thanks and sorry for the mistype
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  4. #4
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    Re: Exponential equation help

    To express the equation in plain text, you could write:

    5^(2x - 1) + 5^x = 10

    To use $\displaystyle \LaTeX$ wrap the following code:

    5^{2x-1}+5^x=10

    In TEX tags, which you can generate using the $\displaystyle \Sigma$ button on the toolbar. This will give you:

    $\displaystyle 5^{2x-1}+5^x=10$

    For this equation, I would use a bit of trial and error. Let's let $\displaystyle x=0$:

    $\displaystyle 5^{2(0)-1}+5^0=\frac{1}{5}+1=1.2<10$

    So, we see $\displaystyle 0<x$. Next try $\displaystyle x=1$...what do you find?
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    Re: Exponential equation help

    Nothing. Couldn't figure it out. I'm still trying.
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    Re: Exponential equation help

    I think I got it. X=1
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: Exponential equation help

    Quote Originally Posted by Nfalconer View Post
    I think I got it. X=1
    Yes, when x = 1, we get:

    5 + 5 = 10

    and this is true.
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  8. #8
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    Re: Exponential equation help

    What we should really do here, and I apologize for being thick, is take the original equation and multiply through by 5 to get:

    $\displaystyle 5^{2x}+5\cdot5^x=50$

    Write as quadratic in $\displaystyle 5^x$ in standard form:

    $\displaystyle \left(5^x\right)^2+5\cdot5^x-50=0$

    Factor:

    $\displaystyle \left(5^x+10\right)\left(5^x-5\right)=0$

    As $\displaystyle 0<5^x$ for all real $\displaystyle x$, we are left with:

    $\displaystyle 5^x-5=0$

    $\displaystyle 5^x=5^1\implies x=1$
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    Re: Exponential equation help

    Quote Originally Posted by MarkFL View Post

    $\displaystyle \left(5^x\right)^2+5\cdot5^x-50=0$
    Or, the original poster referenced substitution. Let y = $\displaystyle \ 5^x $.
    Then factor $\displaystyle \ y^2 - 5y - 50 = 0$.

    Solve that for y. And then substitute back for x, noting the allowable real value of x.
    Last edited by greg1313; Jul 8th 2018 at 08:57 AM.
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    Re: Exponential equation help

    I don't understand the substitution way of solving the equation. Could you elaborate?
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  11. #11
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    Re: Exponential equation help

    Quote Originally Posted by Nfalconer View Post
    I don't understand the substitution way of solving the equation. Could you elaborate?
    $5^{2x-1}+5^x=10$ let $y=5^x$ then we get $(5^x)^2(5^{-1})+5^x-10=0$ so that
    $ \begin{align*}(5^{-1})y^{2}+y-10&=0 \\y^2+5y-50&=0\\(y+10)(y-5)&=0\\(y+10)~&\wedge~(y-5)=0\\y=-10 ~&\wedge~y=5\end{align*}$

    From that we get $5^x=-10~\wedge~5^x=5$
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