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Math Help - solve for log equation

  1. #1
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    solve for log equation

    log3x + log3(x + 2) = 1

    how do i prove this?

    So far, i have:
    = log3[x(x+2)]
    = log3[x^2 + 2x]

    and i can't get any farther?
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  2. #2
    RRH
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    Quote Originally Posted by snypeshow View Post
    log3x + log3(x + 2) = 1

    how do i prove this?

    So far, i have:
    = log3[x(x+2)]
    = log3[x^2 + 2x]

    and i can't get any farther?
    Change is from a log to a non log form; therefore, it will now look like this:

    <br />
x^2+2x=3^1<br />

    Can you solve it now? In addition, don't forget to check your answers
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  3. #3
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    so I would get x = sqrt(3 - 2x)

    ?
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  4. #4
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    Yes, because to get rid of the power u need to square root the rest
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  5. #5
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    YES! thanks alot guys!!!!

    man, there's nothing you guys cant answer :P
    Last edited by snypeshow; November 22nd 2009 at 04:41 PM.
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  6. #6
    RRH
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    Quote Originally Posted by snypeshow View Post
    so I would get x = sqrt(3 - 2x)

    ?
    I do not agree with your answer ...

    <br />
x^2+2x=3^1<br />

    move the subtract 3 from both sides

    <br />
x^2+2x-3=0<br />

    factor

    <br />
(x+3)(x-1)=0<br />

    solve for x:

    x=-3, x=1

    check to make sure you do not get a negative log .... since -3 does give you a negative log

    your solution set is 1
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  7. #7
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    Quote Originally Posted by RRH View Post
    I do not agree with your answer ...

    <br />
x^2+2x=3^1<br />

    move the subtract 3 from both sides

    <br />
x^2+2x-3=0<br />

    factor

    <br />
(x+3)(x-1)=0<br />

    solve for x:

    x=-3, x=1

    check to make sure you do not get a negative log .... since -3 does give you a negative log

    your solution set is 1
    Yeh you are right! Hmpff made a silly mistake there. No point in calculating X at one side if there's still a X at the other side
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