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Math Help - quick question on product rule and equality rule for logs

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    quick question on product rule and equality rule for logs

    so I have an equation:

    ln (2x+1) + ln(x-2) -2lnx = 0

    so to be able to use product rule I added the 2lnx and cancelled out for equality rule

    ln(2x+1)+ln(x-2) = 2 ln x

    then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

    (2x+1)(x-2) = 2x.

    Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks
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    Re: quick question on product rule and equality rule for logs

    Quote Originally Posted by RYdis View Post
    so I have an equation:

    ln (2x+1) + ln(x-2) -2lnx = 0

    so to be able to use product rule I added the 2lnx and cancelled out for equality rule

    ln(2x+1)+ln(x-2) = 2 ln x

    then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

    (2x+1)(x-2) = 2x.

    Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks

    Nope. You cannot simply cancel ln(x) while that 2 is infront of it. Your working is good up to \ln((2x+1)(x-2)) = 2\ln(x)

    Your next step should be to use the log power rule on the RHS
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    Re: quick question on product rule and equality rule for logs

    Hmm..still having difficulties..guess nit understanding this part on my own
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    Re: quick question on product rule and equality rule for logs

    Quote Originally Posted by RYdis View Post
    so I have an equation:

    ln (2x+1) + ln(x-2) -2lnx = 0

    so to be able to use product rule I added the 2lnx and cancelled out for equality rule

    ln(2x+1)+ln(x-2) = 2 ln x

    then ln((2x+1)(x-2)) = 2 ln x, can cancel out the ln so we have

    (2x+1)(x-2) = 2x.

    Im little confused on what to do next. Should I foil out the left side then add the right? Think im forgetting a step here to fully solve this. Thanks
    \displaystyle \begin{align*} \ln{(2x+1)} + \ln{(x - 2)} - 2\ln{(x)} &= 0 \\ \ln{(2x + 1)} + \ln{(x-2)} - \ln{\left(x^2\right)} &= 0 \\ \ln{[(2x+1)(x-2)]} - \ln{\left(x^2\right)} &= 0 \\ \ln{\left[\frac{(2x+1)(x-2)}{x^2}\right]} &= 0 \\ \frac{(2x+1)(x-2)}{x^2} &= e^0 \\ \frac{(2x+1)(x-2)}{x^2} &= 1 \\ (2x+1)(x-2) &= x^2 \end{align*}

    Go from here...
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    Re: quick question on product rule and equality rule for logs

    actually had a slight typo in the problem. But here we go

    going off what you had this is what we actually see (x-3 not x-2)

    (2x+1)(x-3) = x^2..can we divide out the x's?

    [(2x+1)(x-2)]/x^2 = [(2+1)(-3)]/x = 0
    (2+1)(-3) = x
    x=-9?

    everything I know tells me this should be wrong..since x^2 usually means we should have two answers.

    However I was unsure on how to factor it..
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    Re: quick question on product rule and equality rule for logs

    Quote Originally Posted by RYdis View Post
    actually had a slight typo in the problem. But here we go

    going off what you had this is what we actually see (x-3 not x-2)

    (2x+1)(x-3) = x^2..can we divide out the x's?

    [(2x+1)(x-2)]/x^2 = [(2+1)(-3)]/x = 0
    (2+1)(-3) = x
    x=-9?

    everything I know tells me this should be wrong..since x^2 usually means we should have two answers.

    However I was unsure on how to factor it..
    Yes you are wrong in thinking you should divide by x^2. Expand the brackets, move everything to one side, and factorise.
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    Re: quick question on product rule and equality rule for logs

    yes I see, so we get

    2x^2-6x+x-3 = x^2
    2x^2-5x-3 = x^2
    2x^2-x^2-5x-3 = x^2-5x-3 = 0

    this doesn't factor even so x= +- (5+sqrt(37))/2

    trying to relearn this on your own is rather difficult lol :/
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    Re: quick question on product rule and equality rule for logs

    Quote Originally Posted by RYdis View Post
    yes I see, so we get

    2x^2-6x+x-3 = x^2
    2x^2-5x-3 = x^2
    2x^2-x^2-5x-3 = x^2-5x-3 = 0

    this doesn't factor even so x= +- (5+sqrt(37))/2

    trying to relearn this on your own is rather difficult lol :/
    I'm sure you mean \displaystyle x = \frac{5 \pm \sqrt{37}}{2}... Which is correct
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    Re: quick question on product rule and equality rule for logs

    yes sir. Thank you very much... complicated lol
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