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Turning to a single logarithm then simply.

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Write expression as a logarithm of a single quantity and then simplify if possible.

    (1/4)[log (x²-1)-log (x+1)]+3log x


    2. Relevant equations



    3. The attempt at a solution
    I got (1/4)log(x-1) + 3log x so far
     
  2. jcsd
  3. Nov 4, 2008 #2

    gabbagabbahey

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    [tex]a*log(b)=log(b^a)[/tex]....does that help?
     
  4. Nov 4, 2008 #3
    Not really. i need to know the steps towards the answer.
     
  5. Nov 4, 2008 #4

    jacksonpeeble

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    Remember your properties of logarithms. Multiplication can be converted into an exponent in a different format. There are other posts on this topic.
     
  6. Nov 5, 2008 #5

    gabbagabbahey

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    Well, what does your first term become when you use the property in my previous post with [itex]a=\frac{1}{4}[/itex] and [itex]b=(x^2-1)[/itex]?
     
  7. Nov 5, 2008 #6
    log((x^2-1)^(1/4))?
     
  8. Nov 5, 2008 #7

    gabbagabbahey

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    There was a typo in my last post, I meant [itex]b=(x-1)[/itex] (without the squared!), which would give you:

    [tex]\frac{1}{4} Log(x-1)=Log((x-1)^{1\over{4}})= Log(\sqrt[4]{x-1})[/tex]

    What do you get for your second term using the same rule (with a different [itex]a[/itex] and [itex]b[/itex] this time of course)?
     
  9. Nov 5, 2008 #8
    Can you just show me the steps because you were missing something? Meaning that the 3log x is also a part of the problem.
     
  10. Nov 5, 2008 #9

    gabbagabbahey

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    The 3log(x) is your other term....

    I'm not going to just show you the steps, it is important for your future that you learn to think through these problems on your own.

    Using the rule in my first reply, what can you express 3log(x) as?
     
  11. Nov 5, 2008 #10
    log (x^3)?
     
  12. Nov 5, 2008 #11

    gabbagabbahey

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    Yes, good.

    Now if [itex]\frac{1}{4} Log(x-1)= log(\sqrt[4]{x-1})[/itex] and [itex]3log(x)=log(x^3)[/itex], what is [itex]\frac{1}{4} Log(x-1)+3log(x)[/itex]?
     
  13. Nov 5, 2008 #12
    log((x-1)^(1/4))(x^3)?
     
  14. Nov 5, 2008 #13

    gabbagabbahey

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    Do you mean [itex]log(x^3 \sqrt[4]{x-1})[/itex] or [itex]x^3log(\sqrt[4]{x-1})[/itex]?
     
  15. Nov 5, 2008 #14
    [itex]log(x^3 \sqrt[4]{x-1})[/itex]?
     
  16. Nov 5, 2008 #15

    gabbagabbahey

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    Looks good to me.

    ...If your not sure about your result, try plugging a few random numbers for x into your starting expression, and final expression (with the help of a calculator); you should find that they both give the same value for a given x.
     
  17. Nov 5, 2008 #16
    O k...but how you simplify it?
     
  18. Nov 5, 2008 #17

    gabbagabbahey

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    That's as far as you can simplify it....you have gone from having an expression with 3 terms, to an expression with only 1 term, so you have simplified it a fair bit.
     
  19. Nov 5, 2008 #18
    Don't get it.
     
  20. Nov 5, 2008 #19

    gabbagabbahey

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    [itex]log(x^3 \sqrt[4]{x-1})[/itex] looks to me like a more simple expression than [itex]\frac{1}{4}[log (x^2-1)-log (x+1)]+3log(x)[/itex]....would you not agree?

    If so, then you've certainly simplified it at least a little. And I don't see any way to simplify it any more than that....do you?

    If not, then your done: [itex]log(x^3 \sqrt[4]{x-1})[/itex] is your final answer!

    Do you follow?
     
  21. Nov 5, 2008 #20
    Ohh...I get it now.
     
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