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Math Help - logarithmic equation

  1. #1
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    logarithmic equation

    logarithmic equation-log.jpg
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  2. #2
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    Re: logarithmic equation

    \displaystyle \begin{align*} \frac{1}{2}\log{(5 + 7x)} - \log{9} + \log{2} &= 1 - \frac{1}{2}\log{(7 + 2x)} \\ \frac{1}{2}\log{(5 + 7x)} + \frac{1}{2}\log{(7 + 2x)} + \log{\frac{2}{9}} &= 1\\ \frac{1}{2}\log{\left[ (5 + 7x)(7 + 2x) \right]} + \log{\frac{2}{9}} &= 1 \\ \log{\left\{ \left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} \right\}} + \log{\frac{2}{9}} &= 1 \\ \log{\left\{ \frac{2}{9}\left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} \right\}} &= 1 \\ \frac{2}{9}\left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} &= e^1 \\ \left[ (5 + 7x)(7 + 2x) \right]^{\frac{1}{2}} &= \frac{9e}{2} \\ (5 + 7x)(7 + 2x) &= \frac{81e^2}{4} \\ 35 + 10x + 49x + 14x^2 &= \frac{81e^2}{4} \\ 14x^2 + 59x + 35 + \frac{81e^2}{4} &= 0 \end{align*}

    You can now solve this using the Quadratic Formula.
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    Re: logarithmic equation

    Can you do quadratic aswell - I havent done any with e2
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    Re: logarithmic equation

    Quote Originally Posted by skaters View Post
    Can you do quadratic aswell - I havent done any with e2
    Yes I can, but I won't. It works exactly the same way, your c value just happens to be \displaystyle \begin{align*} 35 + \frac{81e^2}{4} \end{align*}.
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  5. #5
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    Re: logarithmic equation

    Hello, skaters!

    I assume that those logs are all base-ten.


    \tfrac{1}{2}\log{(5 + 7x)} - \log{9} + \log{2} \:=\: 1 - \tfrac{1}{2}\log{(7 + 2x)}

    \begin{array}{cc}\text{We have:} & \tfrac{1}{2}\log{(7x+5)} + \tfrac{1}{2}\log{(2x+7)} + \log{\tfrac{2}{9}} \:=\: 1 \\ \\ & \log(7x+5)^{\frac{1}{2}} + \log(2x +7)^{\frac{1}{2}} + \log(\tfrac{2}{9}) \:=\:1 \\ \\ \text{Combine:} & \log\left[\tfrac{2}{9}(7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2  }}\right] \:=\:\log(10) \\ \\ \text{"Un-log":} & \tfrac{2}{9}(7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2  }} \;=\;10 \\ \\ & (7x+5)^{\frac{1}{2}}(2x+7)^{\frac{1}{2}} \;=\;45 \\ \\ \text{Square:} & (7x + 5)(2x+7) \;=\;2025 \\ \\ & 14x^2 + 59x + 25 \;=\;2025 \\ \\ & 14x^2 + 59x - 1990 \;=\;0 \\ \\ \text{Factor:} & (x-10)(14x-199) \;=\;0 \\ \\ \text{Hence:} & x \;=\;10,\,\text{-}\frac{199}{14} \end{array}


    We must discard the negative root.

    Therefore: . x\,=\,10
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    Re: logarithmic equation

    Thanks !
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