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Math Help - logarithm of a single quantity

  1. #1
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    logarithm of a single quantity

    This problem is on my summer packet for pre cal honors this year.

    Write the following expression as a logarithm of a single quantity:
    1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2


    I tried my best to solve it and ended up with this:

    ln[(x+1)(x-1)^2(x^3)/(2x+5)^3]^1/2-2

    I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kitty216 View Post
    This problem is on my summer packet for pre cal honors this year.

    Write the following expression as a logarithm of a single quantity:
    1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2


    I tried my best to solve it and ended up with this:

    ln[(x+1)(x-1)^2(x^3)/(2x+5)^3]^1/2-2

    I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]
    almost! you forgot the guy in red
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  3. #3
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    o.o oh.. it turned into (x^3) or is that incorrect?

    thankies!!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Kitty216 View Post
    This problem is on my summer packet for pre cal honors this year.

    Write the following expression as a logarithm of a single quantity:
    1/2[ln(x+1)+2ln(x-1)-3ln(2x+5)]+3lnx-2


    I tried my best to solve it and ended up with this:

    ln[(x+1)(x-1)^2(x^3)/(2x+5)^3]^1/2-2

    I'm not sure if the two at the end can sit out like that. It would be great if someone can help =]
    \tfrac{1}{2}\left[\ln(x+1)+2\ln(x-1)-3\ln(2x+5)\right]+3\ln(x)-2

    Distribute the \tfrac{1}{2} throughout...

    \tfrac{1}{2}\ln(x+1)+\ln(x-1)-\tfrac{3}{2}\ln(2x+5)+3\ln(x)-2

    Use the power rule of logs...

    \ln(\sqrt{x+1})+\ln(x-1)-\ln((2x+5)^{\tfrac{3}{2}})+\ln(x^3)-2

    Now apply the sum and difference properties of logs:

    \ln(\sqrt{x+1}(x-1)x^3)-\ln(2x+5)^{\frac{3}{2}}-2

    \implies\ln\left(\frac{\sqrt{x+1}(x-1)x^3}{(2x+5)^{\frac{3}{2}}}\right)-2

    Now how do we bring the 2 into the logarithm?

    We know that \ln(e)=1, thus 2\ln(e)=2.

    Therefore, we now have:

    \ln\left(\frac{\sqrt{x+1}(x-1)x^3}{(2x+5)^{\frac{3}{2}}}\right)-2\ln(e)

    \implies\color{red}\boxed{\ln\left(\frac{\sqrt{x+1  }(x-1)x^3}{e^2(2x+5)^{\frac{3}{2}}}\right)}

    Does this make sense?

    I hope this clarifies things!!

    --Chris

    EDIT: Jhevon beat me...
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  5. #5
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    WOW! I get it! I would never have thought about putting in the ln(e).

    Thank you guys!!
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  6. #6
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    logaritem of a single quantity

    Hi ,this is Kevin i had completed my Mathematics.and the concepts are very interested.This may be help full for the math students And for the young youth.And they may learn some knowledge from this forum.And i really appreciate for this thought.
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    Kevin00

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