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Thread: extraneous roots with logs?

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    extraneous roots with logs?

    Hi, I have these four questions that I cannot seem to understand what they are asking me to solve. I do not know how to begin the question.

    The questions are as followed:

    3. Solve. Identify and reject any extraneous roots. Check your solutions using graphing technology.

    a) $\displaystyle log x + log (x-4) = 1$

    b) $\displaystyle log x^3 - log 2 = log (2x^2)$

    c) $\displaystyle log (v-1) = 2 + log (v-16)$

    d) $\displaystyle 1 + log y = log (y+9)$

    How would I do these? My teacher only showed us the way to solve these when they are adding?
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    Quote Originally Posted by Barthayn View Post
    Hi, I have these four questions that I cannot seem to understand what they are asking me to solve. I do not know how to begin the question.

    The questions are as followed:

    3. Solve. Identify and reject any extraneous roots. Check your solutions using graphing technology.

    a) $\displaystyle log x + log (x-4) = 1$

    b) $\displaystyle log x^3 - log 2 = log (2x^2)$

    c) $\displaystyle log (v-1) = 2 + log (v-16)$

    d) $\displaystyle 1 + log y = log (y+9)$

    How would I do these? My teacher only showed us the way to solve these when they are adding?
    I assume these are natural logarithms? As in, base $\displaystyle e$?

    1. $\displaystyle \log{x} + \log{(x + 4)} = 1$

    $\displaystyle \log{[x(x + 4)]} = 1$

    $\displaystyle x(x + 4) = e^1$

    $\displaystyle x(x + 4) = e$

    $\displaystyle x^2 + 4x = e$

    $\displaystyle x^2 + 4x - e = 0$

    Now use the Quadratic Formula.


    2. $\displaystyle \log (x^3) - \log 2 = \log (2x^2)$

    $\displaystyle \log{\frac{x^3}{2}} = \log{(2x^2)}$

    $\displaystyle \frac{x^3}{2} = 2x^2$

    $\displaystyle \frac{1}{3}x^3 - 2x^2 = 0$

    $\displaystyle x^2\left(\frac{1}{3}x - 2\right) = 0$

    Solve for $\displaystyle x$.


    3. $\displaystyle \log (v-1) = 2 + \log (v-16)$

    $\displaystyle \log{(v - 1)} - \log{(v - 16)} = 2$

    $\displaystyle \log{\left(\frac{v - 1}{v - 16}\right)} = 2$

    $\displaystyle \frac{v - 1}{v - 16} = e^2$

    $\displaystyle \frac{v - 16 + 15}{v - 16} = e^2$

    $\displaystyle 1 + \frac{15}{v - 16} = e^2$

    $\displaystyle \frac{15}{v - 16} = e^2 - 1$

    $\displaystyle \frac{v - 16}{15} = \frac{1}{e^2 - 1}$

    $\displaystyle v - 16 = \frac{15}{e^2 - 1}$

    $\displaystyle v = \frac{15}{e^2 - 1} + 16$.


    4. $\displaystyle 1 + \log y = \log (y+9)$

    $\displaystyle \log{(y + 9)} - \log{y} = 1$

    $\displaystyle \log{\frac{y + 9}{y}} = 1$

    $\displaystyle \frac{y + 9}{y} = e^1$

    $\displaystyle \frac{y + 9}{y} = e$

    $\displaystyle 1 + \frac{9}{y} = e$

    $\displaystyle \frac{9}{y} = e - 1$

    $\displaystyle \frac{y}{9} = \frac{1}{e - 1}$

    $\displaystyle y = \frac{9}{e - 1}$.
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