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Homework Help: Isolating a function variable

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Im supposed to calculate the value of x for

    20*log10([itex]\frac{\sqrt{x^2 +1}}{cos(x)}[/itex]) = 3




    3. The attempt at a solution
    I start off by dividing by 20 on both sides so I just have the expression with log.. and my idea henceforth is to take the inverse of cosine to get rid of cos(x), and square the square root to cancel out.. but the logarithm part has me stuck.. and im not sure how to get rid of/move it.. ive thought about multiplying with its inverse, but that has not gotten me any results.. any tips?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 30, 2012
  2. jcsd
  3. Sep 30, 2012 #2
    How about [itex]e[/itex]?

    But I can't really tell what to do after that.
     
  4. Sep 30, 2012 #3

    SammyS

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    You edited your original post!

    Now, there is no equation!

    (Formerly it said 20*log[itex]\frac{\sqrt{x^2 +1}}{\cos(x)}=3\ .[/itex])
     
  5. Sep 30, 2012 #4
    woops, I edited log to log10.. and suddenly the equation was gone :b
     
  6. Sep 30, 2012 #5
    I think propertieds of logs are important here.

    log10 u/v = log10(u)-log10(v)

    So, your original equation can be rewritten as:
    20(log10(u)-log10(v))=3
    with the numerator and denominator substituted for u and v, respectively. (hopefuly that makes sense)

    Does that help?
     
  7. Sep 30, 2012 #6
    Possibly, ill try
     
  8. Sep 30, 2012 #7
    Not really.. I end up with a slightly prettier equation, but im not sure how to isolate x from log(cos(x)
     
  9. Sep 30, 2012 #8
    ah, well, there's a trig identity to help with that, I'm sure...
    Off of the top of my head, there's always cos^2(x)+sin^2(x)=1, so...
    I'll have to look some stuff up and see if it'll help.
     
  10. Sep 30, 2012 #9

    SammyS

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    If [itex]\log_{10}(A)=B\,,[/itex] then [itex]A=10^B\ .[/itex]
     
  11. Sep 30, 2012 #10
    All I see right now is something like this (where I'm using 10 instead of e) [itex]\sqrt{x^2 + 1} = a \cos{x}[/itex] where [itex]a = 10^{3/20}[/itex] then use the trig identity ohms law mentioned. (But multiply the identity by -1)
     
  12. Sep 30, 2012 #11
    Hmm.. so far I have at least gotten rid of the log part, and I now have:

    [itex]\frac{\sqrt{x^2 +1}}{cos(x)}[/itex] = 10[itex]\frac{3}{20}[/itex]

    Which I can rearrange a bit.. but that cos(x) is causing me troubles!
     
  13. Sep 30, 2012 #12
    fyi, for the numerator portion, I simplified it to 1/3 log(x)
     
  14. Sep 30, 2012 #13

    SammyS

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    I do believe that solving this requires using a numerical method.
     
  15. Sep 30, 2012 #14
    1/cos(x) would = sec(x)...
    so, log(1/cos(x)) = log(1) - log(sec(x))
    since log(1) = 0, we can just get rid of that

    just not sure how much that helps
     
  16. Sep 30, 2012 #15
    so, you're saying just plug and puke?
    :)
     
  17. Sep 30, 2012 #16

    SammyS

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    I've reviewed my posts, and no, I didn't say that. (LOL !)

    OP may want to put his equation into a more suitable form before trying a numerical method.

    Also, a graph may help in choosing a starting approximation.
     
    Last edited: Sep 30, 2012
  18. Sep 30, 2012 #17
    I missed this earlier, but... I'm not sure what you mean with that equality. It's not true as stated (see graph):
    Isolating a function variable graph.jpg
     
  19. Sep 30, 2012 #18
    I guess that I'm just not clear what you mean by "requires using a numerical method".
    I feel like I should know what you mean, but... :blushing:
     
  20. Sep 30, 2012 #19
    I dont mean anything in particular, just me trying to isolate x in the equation
     
  21. Sep 30, 2012 #20

    SammyS

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    Actually, that equality is true !

    It leads to [itex]\displaystyle \sqrt{x^2+1}=10^{3/20}\cos(x)\ .[/itex]

    Tyrone123,
    Notice that [itex]\left(10^{3/20}\right)^2\approx1.99526\,,[/itex] which is very nearly 2, so that [itex]10^{3/20}[/itex] is just a little less that [itex]\sqrt{2}\ .[/itex]

    Also, it may help to square that above equation, which then becomes:
    [itex]\displaystyle x^2 +1=10^{3/10}\cos^2(x)\ .[/itex]​

    For small values of x, a good approximation for cos(x) is:
    [itex]\displaystyle
    \cos(x)\approx 1-\frac{x^2}{2}\ . [/itex]​

    Plug that into the previous equation and solve for x to get a good starting place for any numerical method you choose.
     
  22. Sep 30, 2012 #21
    man, trig makes my head hurt... :grumpy:
    I don't get it. I'd drop it, but my sense is that this is part of the problem that Tyrone is having too (maybe I'm wrong, but...).

    I know that graphs don't prove anything, but the graphs are (extremely) different (even with the cos(x) product added to the one side), so... I'm confused.
     
  23. Sep 30, 2012 #22

    SammyS

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    Here are the graphs of [itex]y=\sqrt{x^2+1}[/itex] and [itex]y=10^{3/20}\cos(x)[/itex] overlaid on the same graph (from WolframAlpha).

    attachment.php?attachmentid=51362&stc=1&d=1349039177.gif
     

    Attached Files:

  24. Sep 30, 2012 #23

    SammyS

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    What is graphed here?

    attachment.php?attachmentid=51351&d=1349029962.jpg
     
  25. Sep 30, 2012 #24

    SammyS

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    Take Tyrone's original equation.
    [itex]\displaystyle 20\,\log_{10}\left(\frac{\sqrt{x^2 +1}}{\cos(x)}\right)=3\ [/itex]​

    Do a little basic algebra & apply some of the log rules you suggested:
    [itex] \displaystyle \log(\sqrt{x^2+1})-\log(\cos(x))=\frac{3}{20} [/itex]

    [itex] \displaystyle \log(\sqrt{x^2+1})=\frac{3}{20}+\log(\cos(x)) [/itex]

    If you use each side as the exponent with a base of 10, you get:
    [itex]\displaystyle \sqrt{x^2+1}=10^{3/20}\cos(x)\ [/itex]​
     
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