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Thread: logarithms equation

  1. #1
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    logarithms equation

    The curves $\displaystyle y = \frac{1}{3}^x $ and $\displaystyle 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 $\displaystyle \sqrt{2} $

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

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

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

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

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

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

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

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

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



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

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

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

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


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

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