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Thread: [SOLVED] Log Help

  1. #1
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    [SOLVED] Log Help

    Hi,

    I need help solving a log equation for x algebraically:

    This is the equation: 2log2(x-6) - log2x = 3

    Help is much appreciated!This is what I have so far:
    2log2(x-6) - log2x = 3 (both logs are base 2)

    log2(x-6)^2 - log2x = 3

    Im not sure what the next step I should take to solve is?

    Any thoughts?

    Thanks!

    Jim.
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  2. #2
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    Hello, Jim!

    $\displaystyle 2\log_2(x-6) - \log_2x \:= \:3$

    This is what I have so far: .$\displaystyle \log_2(x-6)^2 - \log_2x \:= \:3$
    Keep going . . .

    $\displaystyle \log_2\left[\frac{(x-6)^2}{x}\right] \:=\:3 \quad\Rightarrow\quad \frac{(x-6)^2}{x} \:=\:2^3 \quad\Rightarrow\quad (x-6)^2 \:=\:8x$


    And we have: .$\displaystyle x^2-12x + 36 \:=\:8x \quad\Rightarrow\quad x^2 - 20x + 36\:=\:0$

    . . which factors: .$\displaystyle (x-2)(x-18) \:=\:0$

    . . and has roots: .$\displaystyle x \:=\:2,\:18$


    But $\displaystyle x=2$ is an extraneous root.

    The only solution is: .$\displaystyle x \,= \,18$

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  3. #3
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    Quote Originally Posted by jimstewart_NC View Post
    Hi,

    I need help solving a log equation for x algebraically:

    This is the equation: 2log2(x-6) - log2x = 3

    Help is much appreciated!This is what I have so far:
    2log2(x-6) - log2x = 3 (both logs are base 2)

    log2(x-6)^2 - log2x = 3

    Im not sure what the next step I should take to solve is?

    Any thoughts?

    Thanks!

    Jim.
    Both sides of the equation contain logarithms. Use the appropriate base to get rid of the logarithms:

    $\displaystyle 2^{2\log_2(x-6)-\log_(x)} = 2^3$

    $\displaystyle \dfrac{(x-6)^2}{x} = 8~\implies~x^2-12x+36=8x$

    This is a quadratic equation. Solve for x. I've got: $\displaystyle x = 18~\vee~x = 2$

    Since x must be greater or equal 6 the only solution is x = 18


    EDIT: Too fast for me - again!
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