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Condensing an expression

  1. Apr 10, 2007 #1
    i have to condense this expression:

    3/2ln5t^6-3/4lnt^4

    i may have done this in the wrong order but so far i've gotten to this point:
    (3/2ln5t^6)/(3/4lnt^4)


    am i doing this right? is that all i have to do?
     
  2. jcsd
  3. Apr 10, 2007 #2

    cristo

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    Remember the rule: logxa=alogx
     
  4. Apr 10, 2007 #3
    so what do i do about those fractions and the exponents that are already there?
     
  5. Apr 10, 2007 #4

    cristo

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    Well, start from the top again, [since you use the incorrect division of logs rule; the correct one is that loga-logb=log(a/b)].

    Firstly, tidy up the expression and remove the exponents, so, what would [tex]\frac{3}{2}\ln 5t^6[/tex] simplify to, using the rule I gave in post 2? Similarly, what would the second term become?

    Then use the division property I give above to combine the two into one logarithm.
     
  6. Apr 10, 2007 #5

    HallsofIvy

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    What exactly do you mean by "condensing" the expression? Writing it as a single expression. Your basic idea is right but you are misremembering a "law of logarithms".
    ln a- ln b= ln (a/b), NOT "ln(a)/ln(b)". Also, as cristo pointed out, you need to handle those coefficients, (3/2) and (3/4).
    [tex]\frac{3}{2}ln(5t^6)= ln((5t^6)^{3/2}[/tex]
     
  7. Apr 10, 2007 #6
    okay so i got this:

    (ln5t^9)/(lnt^3)

    i still don't feel like i did that right =/
     
  8. Apr 10, 2007 #7

    cristo

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    Sorry, I read that as ln(5t)^6 and then used horrible notation in my post!

    You're still using the law of subtraction of logs incorrectly! Also, looking at Halls' post, note that your first term should be [tex]\ln(5^{3/2}t^9)[/tex]. Then look again at the rule loga-logb=log(a/b)
     
  9. Apr 10, 2007 #8
    okay i see how i was doing it wrong now...sorry had a dumb moment.

    so then would i get
    ln(5^3/2*t^9)/t^3) ?
     
  10. Apr 10, 2007 #9

    cristo

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    Ok, can this be simplified?

    [Hint: [tex]\frac{x^a}{x^b}=x^{a-b}[/tex] ]
     
  11. Apr 10, 2007 #10
    ln(5^3/2)t^6 ?
     
  12. Apr 10, 2007 #11

    cristo

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    Yea, it's all in the argument of the log function though: [tex]\ln(5^{3/2}t^6)[/tex]
     
  13. Apr 10, 2007 #12
    yay! thank you so much!
     
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