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Math Help - Urgent log functions

  1. #1
    Junior Member
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    Post Urgent log functions

    Hi, i mixed up my log properties a bit and got these problems wrong.

    1. find the inverse of the function y = logbase7(x + 3) + 7 <--- for this one i know that i have to interchange the x and y, but i dont know where to go from there

    2. Solve the following for x and leave in exact form. Check for extraneous roots.

    a. ln(2x - 2) - ln(x-1) = ln(x)

    b. (log(8x) - 2log(2x))/logx = 1

    c. 2ln(3x) + 5 = 8

    d. log(x) = 90 - log(x - 90)

    e. x^2(log(x)) - 4(log(x)) = 0 <--- for this one there is a hint, it says to factor first, but i can't seem to figure it out.

    Any help is much appreciated, thanks a lot.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, leviathanwave!

    1. Find the inverse of: . y \:= \:\log_7(x + 3) + 7

    "Switch" x and y\!:\;\;x \;=\;\log_7(y + 3) + 7

    Solve for y\!:\;\;x - 7 \;=\;\log_7(y + 3)

    . . . . . . . . . y + 3 \;=\;7^{x-7}

    . . . . . . . . . . . . y \;=\;7^{x-7} - 3



    2. Solve the following for x and leave in exact form. Check for extraneous roots.

    Note that \log(x) exists for x > 0
    . . For other values, the log does not exist (DNE).


    a)\;\;\ln(2x - 2) - \ln(x-1) \:= \:\ln(x)
    We have: . \ln\left(\frac{2x-2}{x-1}\right)\:=\:\ln(x)

    "Un-log": . \frac{2x-2}{x-1} \:=\:x\quad\Rightarrow\quad 2x - 2 \;=\;x^2 - x

    We have a quadratic: . x^2 - 3x + 2 \;=\;0

    . . which factors: . (x - 1)(x - 2)\;=\;0

    . . and has roots: . x \:=\:1,\,2


    For x = 1, DNE.

    The only root is: . \boxed{x = 2}



    b)\;\;\frac{\log(8x) - 2\log(2x)}{\log(x)} \:= \:1

    We have: . \log(8x) - \log(2x)^2\;=\;\log(x)\quad\Rightarrow\quad \log(8x) - \log(4x^2) \;=\;\log(x)

    . . \log\left(\frac{8x}{4x^2}\right)\;=\;\log(x)\quad\  Rightarrow\quad \log\left(\frac{2}{x}\right)\;=\;\log(x)

    Un-log: . \frac{2}{x}\;=\;x\quad\Rightarrow\quad x^2 \,= \,2\quad\Rightarrow\quad x \,=\,\pm\sqrt{2}


    For x = -\sqrt{2}, DNE.

    The only root is: . \boxed{x\,=\,\sqrt{2}}



    c)\;\;2\ln(3x) + 5 \:= \:8

    We have: . \ln(3x)^2\;=\;4

    Then: . (3x)^2\;=\;e^3\quad\Rightarrow\quad 3x\;=\;\pm e^{\frac{3}{2}}\quad\Rightarrow\quad x\:=\:\pm\frac{1}{3}e^{\frac{3}{2}}


    For x = -\frac{1}{3}e^{\frac{3}{2}}, DNE.

    The only root is: . \boxed{x\,=\,\frac{1}{3}e^{\frac{3}{2}}}



    d)\;\;\log(x) \:= \:90 - \log(x - 90)

    Is there a typo?. We can't solve this one . . .

    We have: . \log(x) + \log(x-90)\;=\;90\quad\Rightarrow\quad \log\left[x(x - 90)\right]\;=\;90

    Then: . x(x - 90)\;=\;10^{90}\quad\Rightarrow\quad x^2 - 90x - 10^{90} \;=\;0 ??



    e)\;\;x^2\log(x) - 4\log(x) \: = \:0

    Factor: . \log(x)\!\cdot\!(x^2 - 4)\;=\;0

    Factor: . \log(x)\!\cdot\!(x - 2)(x + 2)\;=\;0

    Then: . \begin{Bmatrix}\log(x) = 0& \Rightarrow & x = 1 \\ x-2 \:=\:0 & \Rightarrow & x = 2 \\ x + 2\:=\:0 & \Rightarrow & x = \text{-}2\end{Bmatrix}


    For x = \text{-}2, DNE.

    The roots are: . \boxed{x \:=\:1,\,2}

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