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Math Help - 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: . \dfrac{1}{\log_4\left(\frac{x+1}{x+2}\right)} \;<\; \dfrac{1}{\log_4(x+3)}\;\;[1]

    Note that the domain is: . x > -1


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

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

    Multiply [1] by negative \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 x > -1,\;\log_4(x+3) \:>\:0

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


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

    . . which simplifies to: . 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 x in the domain: . (-1,\:\infty)

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