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Thread: trignometry

  1. #1
    Member grgrsanjay's Avatar
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    Post trignometry

    solve the eqn:

    1/log(base 4) (x+1)/(x+2) < 1/log(base 4) (x+3)

    regret me
    the topic is LOGARITHMS
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  2. #2
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    Hello, grgrsanjay!

    I have a very long solution . . . and it may be incomplete.
    Someone check it carefully . . . please!


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

    We begin with: .$\displaystyle 1 \:<\:2$

    Add $\displaystyle x\!:\;\;x+1 \:<\:x+2\;\;[2]$

    Domain: $\displaystyle x \,>\,-1 \quad\Rightarrow\quad x+1 \,>\,0 \quad\Rightarrow\quad x+2 \,>\,0$

    Divide [2] by positive $\displaystyle x+2\!:$ . $\displaystyle \dfrac{x+1}{x+2} \:<\:1 $

    Take logs, base 4: .$\displaystyle \log_4\!\left(\dfrac{x+1}{x+2}\right)\:<\:\log_4(1 ) \:=\:0$

    That is: .$\displaystyle \log_4\!\left(\dfrac{x+1}{x+2}\right)$ .is negative.


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


    Since $\displaystyle x \,>\, -1$, then $\displaystyle x+3 \,>\,2 \quad\Rightarrow\quad \log_4(x+3) \:>\:0 $


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

    Take anti-logs: .$\displaystyle x+3\;>\;\dfrac{x+1}{x+2} \quad\Rightarrow\quad x^2 + 4x + 5 \;>\;0 $

    This is an up-opening parabola with its vertex at (-2, 1).
    . . Hence, it is always above the $\displaystyle x$-axis.


    The inequality is true for all $\displaystyle x$ in the domain: .$\displaystyle x \:>\:-1$

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