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Math Help - [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!

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

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

    \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: . x^2-12x + 36 \:=\:8x \quad\Rightarrow\quad x^2 - 20x + 36\:=\:0

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

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


    But x=2 is an extraneous root.

    The only solution is: . 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:

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

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

    This is a quadratic equation. Solve for x. I've got: 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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