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Thread: logarithms

  1. #1
    Member grgrsanjay's Avatar
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    logarithms

    solve the eqn:

    1/log(base 4) (x+1)/(x+2) < 1/log(base 4) (x+3)
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  2. #2
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    Hello, grgrsanjay!

    I already solved this somewhere . . . Here is it again.


    Solve: .$\displaystyle \dfrac{1}{\log_4\left(\frac{x+1}{x+2}\right)} \;<\; \dfrac{1}{\log_4(x+3)}\;\;[1]$

    Note that the domain is: .$\displaystyle x > -1$


    For $\displaystyle x > -1,$ .$\displaystyle \dfrac{x+1}{x+2} \:<\:1$

    . . Hence: .$\displaystyle \log_4\!\left(\dfrac{x+1}{x+2}\right) \:<\:0$

    Multiply [1] by negative $\displaystyle \log_4\!\left(\frac{x+1}{x+2}\right)\!:\;\;1 \;>\;\dfrac{\log_4\left(\frac{x+1}{x+2}\right)}{\l og_4(x+3)}\;\;[2]$


    For $\displaystyle x > -1,\;\log_4(x+3) \:>\:0$

    Multiply [2] by positive $\displaystyle \log_4(x+3)\!:\;\;\log_4(x+3) \;>\;\log_4\!\left(\dfrac{x+1}{x+2}\right)$


    Take anti-logs: . $\displaystyle x+3 \,>\;\dfrac{x+1}{x+2} $

    . . which simplifies to: .$\displaystyle x^2 + 4x + 5 \:>\:0\;\;[3]$


    This is an up-opening parabola whose vertex is at (-2,1).
    . . Hence, the function is always positive.


    The inequality is true for any $\displaystyle x$ in the domain: .$\displaystyle (-1,\:\infty)$

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