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How to use logarithm

  1. Sep 8, 2005 #1

    DB

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    my physics teacher told us that for now well be solving the equation of a line such as [tex]y=kx^n[/tex] by trial and error for finding "n" using the multiplicative change in 2 points i.e [tex]x_1,y_1[/tex],[tex]x_2,y_2[/tex].
    what we would do is for example 2 points (13,1) (6, 1.4):
    [tex]n=\frac{y_2}{y_1}=(\frac{x_2}{x_1})^n[/tex]
    [tex]n=\frac{6}{13}=(\frac{1.4}{1})^n[/tex]
    so:
    [tex]~0.46=1.4^?[/tex]
    we'd have to use trial and error....wat id like to know is how to use logarithm to solve this, he said he might teach it but i was wondering if u guys could fill me in on how to solve this using log:
    [tex]~0.46=1.4^n[/tex],
    the answer for n is about -2.3...
    thanks in advance
     
  2. jcsd
  3. Sep 8, 2005 #2
    You need to know the properties of logarithms
    One is log a^x=x log a

    Try looking in your precalc or algebra II book in order to see all the properties.
     
  4. Sep 8, 2005 #3

    DB

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    i have none of those books, im in sec 5 (grade 11)
    plus i dont understand how to use the property u gave me :frown:
     
    Last edited: Sep 8, 2005
  5. Sep 8, 2005 #4
    Suppose [itex]a=e^b[/itex]. Then [itex]a^{x}=e^{bx}[/itex] which implies [itex]\ln{a^{x}}=\ln{e^{bx}}=bx[/itex]. But since b=ln(a), you can substitute which gives [itex]\ln{a^{x}}=x\ln{a}[/itex].

    Try taking the logarithm of both sides of your equation and use this property with some simple algebra. It should work out.
     
  6. Sep 8, 2005 #5

    TD

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    Well the answer isn't exactly -2.3
    If you want to use logarithms, it goes like this:

    [tex]0.46 = 1.4^n \Leftrightarrow \log 0.46 = \log 1.4^n \Leftrightarrow \log 0.46 = n\log 1.4 \Leftrightarrow n = \frac{{\log 0.46}}
    {{\log 1.4}} \approx - 2.307853977[/tex]

    Since I don't know how much you know, just tell us which steps you don't understand to get further details. As said before, we use that [itex]\log a^x = x\log a[/itex].
     
  7. Sep 8, 2005 #6

    DB

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    i still dont get it, maybe if u could use whole numbers as examples?
     
  8. Sep 8, 2005 #7

    TD

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    It goes exactly the same, I always left the numbers so that wouldn't matter. Only in the last step I made an aproximation but that was done with a computer, you can't do it on your own for arbitrary numbers. To have the exact answer, you just leave the logarithm form.
     
  9. Sep 8, 2005 #8

    DB

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    thanks guys i got it now, this will save me alot of guessing :)
     
  10. Sep 8, 2005 #9

    DB

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    just a question, why are we aloud to bring the exponent "n" down in front of the log as in:
    log(0.46)=log(1.4)^n = log(0.46)= n log(1.4)
    and why then can "n" be brought to the left side of the equation without turning negative?
     
  11. Sep 8, 2005 #10
    This was explained by apmcavoy in this thread.
    We're dividing both sides of the equation by log(1.4).
     
  12. Sep 8, 2005 #11

    DB

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    okay thanks hypermorphism im gonna study apmcavoy's thread, just a question about it: is he referring to the constan "e" ~ 2.7183?
     
  13. Sep 8, 2005 #12
    Yep, that's the one. First discovered as the base of the natural logarithm function.
     
  14. Sep 8, 2005 #13

    HallsofIvy

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    Oh, my God, Oh, my God, Oh, my God.

    If you are working with logarithms you should not even be asking a question like that! I wish there were a way of going back in time and killing your algebra teachers! (Unless of course you simply didn't pay attention to them. In which case, I should simply kill you!)

    We do not ever just "bring a number to the left side"! We do the opposite of what ever operation is done to the number we want to isolate.

    IF we have, for example, x+ a= b, then, since a is added to x, we do the opposite: subtract a from both sides so that we get x+a-a= b-a or x= b- a. Yes, in this one case, the number "a" has "changed sign" or "turned negative".

    IF we have, for example, ax= b, then, since x is multiplied by a, we do the opposite: divide by a. ax/a= b/a or x= b/a. There is no "negative" involved in that!

    IF we have, for example, ex= b, then since x is an exponential, we do the opposite: take the logarithm so that log(ex)= log b or
    x= log b. log(ex)= x because that is the way log is defined!!!
    There is no "negative" involved!

    Sorry if I sound over the top here but I am teaching two sections of "remedial algebra" and I spend most of my time cursing high school algebra teachers!
     
  15. Sep 8, 2005 #14

    DB

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    hallsofivy, relax, i know my algebra, to be honest u've really insulted me by "teaching" me wat you did. i simply just thought that an "n" in front of a log (n log(1.4)) ment something other than multiplication until hypermorphism told me about dividing both sides and i saw the algebra automatically, realizing it was multiplication. i've never used a log before in my life neither the constant "e" so i didnt know wat to expect. ive got no reason to curse my algebra teachers, i do pretty good in it.
     
  16. Sep 8, 2005 #15

    DB

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    im currently in 536 math...do u really think i would be in it if i couldnt do the examples you posted? i sure as hell would have trouble with quadratics, proofs and optimization....
    and i really dont care who you want to kill

    by the way its 500 + 30 + 6, just in case you didnt know how those numbers came to be...
     
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