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Write an expression for f3 in terms of f2

  1. Apr 3, 2005 #1

    tony873004

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    [tex]
    f_3 =0.5\left( {f_1 +f_2 } \right)
    \]
    [/tex]

    Write an expression for [tex]f_3[/tex] in terms of [tex]f_2[/tex].

    I take this to mean that I have to eliminate [tex]f_1[/tex].

    So I start by writing an expression for [tex]f_1[/tex] in terms of [tex]f_2[/tex] and [tex]f_3[/tex].

    [tex]
    \[
    f_3 =0.5f_1 +0.5f_2
    \]
    \[
    0.5f_1 +0.5f_2 -f_3 =0
    \]
    \[
    -0.5f_1 =0.5f_2 -f_3
    \]
    \[
    0.5f_1 =f_3 -0.5f_2
    \]
    \[
    f_1 =\frac{f_3 -0.5f_2 }{0.5}
    \]
    \[
    f_1 =2\left( {f_3 -0.5f_2 } \right)
    \]
    \[
    f_1 =2f_3 -f_2
    \]
    Substitute my new expression for f1 into the original formula
    \[
    f_3 =0.5\left( {f_1 +f_2 } \right)
    \]
    \[
    f_3 =0.5\left( {2f_3 -f_2 +f_2 } \right)
    \]
    \[
    f_3 =f_3 -0.5f_2 +0.5f_2
    \]
    And all I've shown is that
    \[
    f_3 =f_3
    [/tex]

    What did I miss? Is this even possible?
     
  2. jcsd
  3. Apr 3, 2005 #2
    You can't eliminate a variable from an equation using the same equation (unless it's a trivial case, like [itex]x = y - y[/itex]) and get a meaningful result!

    You need a second equation to eliminate [itex]f_1[/itex]...
     
  4. Apr 3, 2005 #3

    xanthym

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    You did nothing wrong. You didn't miss anything. But you didn't accomplish anything, either. You cannot substitute an equation back into itself and accomplish anything useful. All you've done is generate the tautology that f3 = f3, something that's perfectly true and perfectly useless!!

    (P.S. - Wasn't there another equation provided for this problem? Normally you'd use this second equation to eliminate f1.)
    ~~
     
    Last edited: Apr 3, 2005
  5. Apr 3, 2005 #4

    tony873004

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    Thanks.

    That was a lot of work to accomplish nothing!!

    The full question is:

    Suppose the eye perceived a star to be midway in brightness between the m1 =1 and m2 = 6 if the flux of the star were midway between the two, i.e., had a flux f3 = ½(f1+f2). Write down an expression for f3 in terms of f2 then use it to compute the magnitude m3 of the star.

    The other equation is
    [tex]m_2-m_1 = -2.5 log(\frac{f_1}{f_2})[/tex]
     
    Last edited: Apr 3, 2005
  6. Apr 3, 2005 #5
    well, you're given [itex]m_2 = 6, \ m_1 = 1[/itex], so you have

    [tex] 5 = -2.5 \ln{\frac{f_1}{f_2}}[/tex]

    [tex] \Longrightarrow -2 = \ln{f_1} - \ln{f_2} \Longrightarrow f_1 = \frac{f_2}{e^2}[/tex]

    so

    [tex]f_3 = \frac{f_1 + f_2}{2} = f_2\left(\frac{e^{-2}+1}{2}\right)[/tex]
     
  7. Apr 3, 2005 #6

    tony873004

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    It looks like you substituted ln for log. Can you do that? And if not, wouldn't that effect the e part of it?

    There's a reason I love science and hate math :tongue2:
     
  8. Apr 3, 2005 #7

    xanthym

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    We eliminate the Log10 by expressing the relationship with:
    f1/f2 = 2.512^(m2 - m1)
    Then:
    f3 = ½(f1 + f2) ----> f1 = 2*f3 - f2
    ::: ⇒ (2*f3 - f2)/f2 = 2.512^(m2 - m1)
    ::: ⇒ f3 = (1/2)*f2*{1 + 2.512^(m2 - m1)}
    ::: ⇒ f3 = (1/2)*f2*{1 + 2.512^(6 - 1)}
    ::: ⇒ f3 = (50.511)*f2

    (Note: Your magnitude equation is NOT correct. It should be:
    [tex] \color{blue} m_2 \ - \ m_1 \ = \ (2.5) \cdot \log_{10}(\frac{f_1}{f_2})[/tex]
    since magnitudes DECREASE with INCREASING flux. This equation is equivalent to the one used above in the solution.)


    ~~
     
    Last edited: Apr 3, 2005
  9. Apr 3, 2005 #8
    well, in math, [itex]\log[/itex] means the same thing as [itex]\ln[/itex] most of the time :wink:

    If you meant logarithm base [itex]10[/itex], just throw in [itex]10[/itex] wherever I have [itex]e.[/itex]
     
  10. Apr 3, 2005 #9

    xanthym

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    .

    TONY:
    Your magnitude equation is NOT correct. It should be:
    [tex] \color{blue} m_2 \ - \ m_1 \ = \ (2.5) \cdot \log_{10}(\frac{f_1}{f_2})[/tex]
    since magnitudes DECREASE with INCREASING flux.

    The equivalent to the above correct equation was used in the MSG #7 solution.


    ~~
     
    Last edited: Apr 3, 2005
  11. Apr 3, 2005 #10

    tony873004

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    Thanks, Data, it's been a while since I've used log or ln for anything other than to pass an algebra test.

    xanthym, isn't
    m1-m2 = - 2.5 log(f1/f2)
    the same as
    m2-m1 = + 2.5 log(f1/f2)

    I don't know why in my class notes I have it written the way I posted it, but in the book it writes it like you show it.

    Thanks, xanthym. I'm trying to reconstruct how you arrived at the first line: f1/f2 = 2.512^(m2-m1)

    [tex]m_1 -m_2 =-2.5\log \left( {\frac{f_1 }{f_2 }} \right)
    \]
    \[
    \frac{m_1 -m_2 }{-2.5}=\log \left( {\frac{f_1 }{f_2 }} \right)
    \]
    \[
    \frac{f_1 }{f_2 }=\log ^{-1}\left( {\frac{1-6}{-2.5}} \right)
    \]
    \[
    \frac{f_1 }{f_2 }=10^{\left( {\frac{1-6}{-2.5}} \right)}
    \]
    \[
    \frac{f_1 }{f_2 }=2.512
    [/tex]

    I'm missing the ^(m2-m1)
     
  12. Apr 3, 2005 #11
    You wrote [itex]m_2 - m_1 = -2.5 \log_{10}(f_1/f_2)[/itex] earlier. If it's really the way that xanthym said, then instead replace [itex]e[/itex] with [itex]\frac{1}{10}[/itex] everywhere in my solution :smile:
     
  13. Apr 3, 2005 #12

    xanthym

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    See RED note above.


    ~~
     
    Last edited: Apr 3, 2005
  14. Apr 3, 2005 #13

    tony873004

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    Thanks, xanthym. I think I see it now. But I have to hang it up for the night. My brain is fried!! Stay tuned. This isn't due till Wednesday, and I think you guys have put me on the right track with 2 different methods.

    Data, looking at your solution, I never realized that log(x/y) = log(x)-log(y). I just made up a few numbers to see if you can really do that, and it works. But I tried the same thing for sin and it didn't work. I Guess this applies to logs only? We probably covered that in math class. But I can do things 1000 times in math class and never get it. But seeing it once in a physics or science problem and I'll remember it forever. Go figure!!

    Thanks both of you!!
     
  15. Apr 3, 2005 #14
    Yes, it is always true. Here's why:

    Recall that [itex]\log_b{x}[/itex] can be regarded as the inverse of [itex]b^x[/itex], ie.

    [tex] y = b^x \Longleftrightarrow \log_b{y} = x.[/tex]

    Now, I'm sure you remember

    [tex]b^{a-c} = \frac{b^a}{b^c}[/tex]

    so clearly

    [tex]b^{\log_b x - \log_b y} = \frac{b^{\log_b x}}{b^{\log_b y}}[/tex]

    but by our definition above, since [itex]\log_b x[/itex] is the inverse of [itex]b^x[/itex], we have

    [tex] b^{\log_b x} = x, \; b^{\log_b y} = y[/tex]

    so from our expression above, we just get

    [tex]b^{\log_b x - \log_b y} = \frac{x}{y}[/tex]

    [tex] \Longrightarrow \log_b b^{\log_b x - \log_b y} = \log_b \frac{x}{y}[/tex]

    but again recalling the inverse fact, we just get

    [tex] \log_b b^{\log_b x - \log_b y} = \log_b x - \log_b y[/tex]

    so

    [tex] \log_b x - \log_b y = \log_b \frac{x}{y}[/tex]

    as I said :smile:
     
    Last edited: Apr 3, 2005
  16. Apr 3, 2005 #15
    and nothing similar works for sines or anything, since they aren't defined as the inverse of an exponential! :rofl:
     
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