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Math Help - 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)  log x + log (x-4) = 1

    b)  log x^3 - log 2 = log (2x^2)

    c)  log (v-1) = 2 + log (v-16)

    d)  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)  log x + log (x-4) = 1

    b)  log x^3 - log 2 = log (2x^2)

    c)  log (v-1) = 2 + log (v-16)

    d)  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 e?

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

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

    x(x + 4) = e^1

    x(x + 4) = e

    x^2 + 4x = e

    x^2 + 4x - e = 0

    Now use the Quadratic Formula.


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

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

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

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

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

    Solve for x.


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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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