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Math Help - logarithms equation

  1. #1
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    logarithms equation

    The curves  y = \frac{1}{3}^x and  y = 2(3^x) intersect at the point P.
    (c) Find the x-coordinate of P to 2 decimal places and show that the y-coordinate
    of P is  \sqrt{2}

     \frac{1}{3}^x = 2 (3^x)

     x = \frac{lg\frac{1}{2}}{2lg3} = -0.32

    How do I show the corresponding 'y' value =  \sqrt{2} ?

     y = 2(3^{-0.32}) this equals the exact value of root 2, but I need to show it using Algebra?

    can anyone show me?

    Thanks!
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  2. #2
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    Hello, Tweety!

    The curves . \begin{Bmatrix}y &=&2\cdot3^x & {\color{blue}[1]} \\ y &=& \left(\tfrac{1}{3}\right)^x & {\color{blue}[2]} \end{Bmatrix} . intersect at point P.

    (c) Find the x-coordinate of P to 2 decimal places
    . . .and show that the y-coordinate of P is \sqrt{2}.

    Equate [1] and [2]: . 2\cdot3^x \:=\:\frac{1}{3^x}

    Multiply by 3^x\!:\quad 2\cdot3^{2x} \:=\:1 \quad\Rightarrow\quad 3^{2x} \:=\:\frac{1}{2} .[3]

    Take logs: . \ln(3^{2x}) \:=\:\ln\left(\tfrac{1}{2}\right) \:=\:\ln(2^{\text{-}1}) \quad\Rightarrow\quad 2x\ln(3)\:=\:-\ln(2)

    . . x \;=\;-\frac{\ln(2)}{2\ln(3)} \;=\;-0.315464877 \;\approx\;\boxed{-0.32}



    At [3], we have: . 3^{2x} \:=\:\frac{1}{2}

    Take logs (base 3): . \log_3(3^{2x}) \:=\:\log_3\left(\frac{1}{2}\right) \:=\:\log_3(2^{-1})

    \text{Then we have: }\;2x\underbrace{\log_3(3)}_{\text{This is 1}} \;=\;-\log_3(2) \quad\Rightarrow\quad 2x \:=\:-\log_3(2)

    . . x \:=\:-\tfrac{1}{2}\log_3(2) \:=\:\log_3\!\left(2^{-\frac{1}{2}}\right) \:=\:\log_3\!\left(\frac{1}{2^{\frac{1}{2}}}\right  ) \quad\Rightarrow\quad x\:=\:\log_3\left(\frac{1}{\sqrt{2}}\right)


    Substitute into [1]: . y \;=\;2\cdot3^{\log_3\left(\frac{1}{\sqrt{2}}\right  )} \;=\;2\cdot\frac{1}{\sqrt{2}} \quad\Rightarrow\quad y \;=\;\sqrt{2}

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