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Logarithmic Functions

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    I have no idea what to do, can someone point me in the right direction?
     
  2. jcsd
  3. Sep 19, 2007 #2
    First thing I would do is change the bases so they are equivalent.
     
  4. Sep 19, 2007 #3
    How would you do that though =/
     
  5. Sep 19, 2007 #4

    rock.freak667

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    Homework Helper

    Change of base formula:

    [tex] log_a b = \frac{log_c b}{log_c a} [/tex] where c is the new base
     
  6. Sep 19, 2007 #5
    kk ty
     
  7. Sep 19, 2007 #6
    The problem.

    I'll write the problem out because your image is a little hard to see.

    If [tex]\log_{2n}(1944)=\log_{n}(486\sqrt{2}), [/tex] determine the value of [tex]n^{3}[/tex]

    Let me know if you have more problems with this, im fresh on log functions.
     
    Last edited: Sep 20, 2007
  8. Sep 19, 2007 #7
    kk sooo log 1944/ log 2n = log 486/2 / log n

    Do the logs cancel?
     
    Last edited: Sep 19, 2007
  9. Sep 19, 2007 #8
    Did you change to base 10? is that why your not placing the bases? and in the log to your right, I suppose thats not 486/2 but [tex]486\sqrt{2}[/tex]?
     
    Last edited: Sep 20, 2007
  10. Sep 19, 2007 #9
    yea it is that but i didnt know how to do the thingie
     
  11. Sep 19, 2007 #10
    do the logs cancel?
     
  12. Sep 19, 2007 #11
    Oh wow... That problem is interesting I actually think there is a better way to solve it.
     
  13. Sep 20, 2007 #12

    HallsofIvy

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    Since we don't know what you mean by "thingie", that tells us nothing!

    You have already been told that you can write the equation as
    [tex]\frac{log(1944)}{log(2n)}= \frac{log(468)}{log(n)}[/tex]
    Do you see that log(2n)= log(n)+ log(2)?
    And that you can rewrite that equation as
    [tex]\frac{1944}{log(n)+ log(2)}= \frac{log(468)}{log(n)}[/tex]

    I haven't specified the base of "log" because it could be anything. Choose base 10 if you like or natural logarithms. In any case, log(1944), log(2), and log(468) are numbers.

    Could you solve an equation like
    [tex]\frac{A}{x+ B}= \frac{C}{x}[/tex]

    and do you see why your equation is of that form?
     
  14. Sep 20, 2007 #13
    Umm, by this don't you mean:
    [tex]\frac{\log1944}{\log(n)+\log(2)}=\frac{log(486\sqrt{2})}{log(n)}[/tex]
     
  15. Sep 20, 2007 #14
    yea thanks but, i was able to find out what to do.
     
  16. Sep 20, 2007 #15

    HallsofIvy

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    Yes, don't know why I dropped the [itex]\sqrt{2}[/itex].
     
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