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Homework Help: Impossible Problem?

  1. Sep 16, 2010 #1
    Impossible Problem?????????

    Can anyone solve this problem.....

    The age of a machine, x, in years, is related to the probability of breakdown, y, by the formula...

    x = 3 + 1n [y / 1 - y]

    Determine the probability of breakdown for 1, 3 and 10 years.

    You need to explain what you did to acheive each out come.
     
  2. jcsd
  3. Sep 16, 2010 #2

    statdad

    User Avatar
    Homework Helper

    Re: Impossible Problem?????????

    In each case you are given a value of [tex] x [/tex] and asked to solve for the corresponding value of [tex] y [/tex]. You need to rearrange the equation you're provided to be able to do that. By the way, parentheses can help you give a clear presentation of your formula:

    x = 3 + ln(y/(1-y))

    is a better way to write your formula if you don't use the latex feature here in this post. written with latex

    [tex]
    x = 3 + \ln \left(\frac y{1-y}\right)
    [/tex]
     
  4. Sep 16, 2010 #3
    Re: Impossible Problem?????????

    Uh no, this is basic algebra, and you should give it an attempt first.
     
  5. Sep 16, 2010 #4

    Mark44

    Staff: Mentor

    Re: Impossible Problem?????????

    It's not at all impossible. What you need to do is to solve the equation for y in terms of x; i.e., find the inverse relationship.

    BTW, it's not 1n; it's ln, short for logarithmus naturalis.

    Also, you need another pair of parentheses in your equation, like so: x = 3 + ln [y / (1 - y)]
     
  6. Sep 16, 2010 #5
    Re: Impossible Problem?????????

    I know i'm supposed to rearrange the equation , but from there on i'm stuck.... can anyone give me a hint or two?
     
  7. Sep 16, 2010 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Impossible Problem?????????

    x is the time in years and y is the probability of break down. To find "the probability of breakdown for 1, 3 and 10 years", set x equal to 1, 3, and then 10 and solve for y in each equation:
    [tex]x= 3+ ln\left(\frac{y}{1- y}\right)[/tex]
    so you need to solve
    [tex]3+ ln\left(\frac{y}{1- y}\right)= 1[/tex]

    [tex]3+ ln\left(\frac{y}{1- y}\right)= 3[/tex]
    and
    [tex]3+ ln\left(\frac{y}{1- y}\right)= 10[/tex]

    I would start, in each equation by subtracting 3 from both sides. And you invert ln(x) by using its inverse function [itex]e^x[/itex], the exponential function: if ln(y)= x, then [itex]y= e^x[/itex].
     
  8. Sep 16, 2010 #7
    Re: Impossible Problem?????????

    Ok, so now I've got e^-2 [y / 1 - y] = -2

    How do i get a value for y?
     
  9. Sep 16, 2010 #8
    Re: Impossible Problem?????????

    What happened to X?

    Subtracting 3 from both sides gives you:

    [tex]x-3 = ln \left( \frac{y}{1-y} \right) [/tex]
     
    Last edited: Sep 17, 2010
  10. Sep 22, 2010 #9
    Re: Impossible Problem?????????

    So here it is........

    The age of a machine, x, in years, is related to the probability of breakdown, y, by the formula...

    x = 3 + 1n [y / 1 - y]

    Determine the probability of breakdown for 1, 3 and 10 years.

    We start off with :- x = 3 + 1n [y / 1 - y]

    Using Age of machine is 1 Year x = 1 , we get...

    1 = 3 + ln [y / 1 - y]...............

    Using Exponential e, we get.... 1 - 3 (and Exchange Exponential e for + ln..

    = e^-2 = [ y / 1 - y ]........ (Exponential (e) -2 = 0.135)

    Now here's the 'Magic'...

    We need to make 'y' the subject so we need to rearrange the equation...

    So the Equation becomes.... y = 0.135 / 1 + 0.135

    Which = y = 0.135 / 1.135 = 0.1189............

    The probability of breakdown for 1 year is................ 0.1189

    Have a go at 3 years and 10 years and see if you get the same Answers as I did...

    The probability of breakdown for 3 years is............... 0.5
    The probability of breakdown for 10 years is............. 0.999
     
  11. Sep 23, 2010 #10
    Re: Impossible Problem?????????

    This is correct. But why not just solve for Y first, then enter the X values as needed?

    You should come up with

    [tex]y = \frac{e^{x-3}}{1+e^{x-3}} = \frac{e^x}{e^3 + e^x}[/tex]
     
  12. Sep 23, 2010 #11
    Re: Impossible Problem?????????

    Yeah, I like your way of thinking, I 'll keep a note of that. Because up until now I didn't know it could be solved that way. Thanks For the Heads Up.
     
  13. Sep 23, 2010 #12
    Re: Impossible Problem?????????

    But looking at your idea......it's another way.... But, I get to thinking my way is a lot Simpler. (Algebra is all about reversing the equasion anyway).

    i.e........

    3x + 4 = 19
    3x = 19 - 4
    ...and so on.

    Thanks for your input
     
  14. Sep 23, 2010 #13
    Re: Impossible Problem?????????

    No problem.

    Of course, what's simple for one person, isn't necessarily simple for another.

    Knowing that this is a homework forum, I'm apt to think that the question you posted might not have been stated exactly as it was given to you. Or, that the instructor may have wanted to see the equation solved for "y" instead.

    Also, If you needed to find the value of y for many different values of x, it would be much easier to enter my version into a calculator or spreadsheet. Then, all you would need to do is enter new values of x to get corresponding values of y. No extra algebra necessary!
     
  15. Sep 23, 2010 #14
    Re: Impossible Problem?????????

    It's all part of my Mechanical Engineering HNC, and I know a few people I'm studying with used MS Excel...... but come on, using a spreadsheet isn't really showing that you can do the Math, is it. It's not getting the grey matter working.

    But like you said it's easier for some than it is others... but like Esther Rantzen used to say, 'That's Life'. Thanks again for your input, Keep up the good work, Cheers.
     
  16. Oct 7, 2010 #15
    Re: Impossible Problem?????????

    Hi, What do I do to rearragne this Equation to make 'I' the subject.....

    144 x 15.71 + I x 15.71 = (144 + I) 14.05
     
  17. Oct 7, 2010 #16
    Re: Impossible Problem?????????

    What's the Difference between Sine and arcsin? Can I still work out an arcsin problem on my calculator using the sin key?
     
  18. Oct 7, 2010 #17
    Re: Impossible Problem?????????

    First of all, you should start a new post, since this is a new, unrelated question.

    Arcsine is the inverse of Sine. It finds the angle given a ratio, whereas Sine finds the ratio given an angle. (The ratio being the length of the side opposite the angle divided by the length of the hypotenuse of a right triangle).

    Most calculators use [itex]sin^{-1}[/tex] for the Arcsine function, and is usually a 2nd function above the "sin" key.
     
  19. Oct 7, 2010 #18

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Impossible Problem?????????

    zgozvrm is correct. Please post new & unrelated questions by starting a new thread.
     
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