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Me !

  1. Sep 14, 2005 #1
    plz help me !!!!!!!!!!!

    plz help me to solve this problem
    find exponential function f(x)=Ca^x with the 2 given points (0,5)(2,5/9)
    thankzzzzzzzzzz..........
     
  2. jcsd
  3. Sep 14, 2005 #2

    LeonhardEuler

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    Alright, [itex]f(x)=Ca^x[/itex], f(0)=5, and f(2)=5/9. So you have two equations and two unkowns: a and C. You can write the first equation as:
    [tex]5=Ca^0[/tex]
    What is a number to the zero power? Answer that and you have C. Now the second equation is:
    [tex]\frac{5}{9}=Ca^2[/tex]
    Remember, you already found C from the last equation, so you just have to find a. Just isolate "a" and take the square root of both sides:
    [tex]a=\sqrt{\frac{5}{9C}}[/tex]
    Then just plug "a" and "C" back into the original equation and you're done.
     
  4. Sep 14, 2005 #3
    Thanks a lot for ur help. Actually i did the problem 5 times already and i knew C=5 but i can not find a=? and My assignment is about to be due in half an hour. Again thanks for ur help
     
  5. Sep 14, 2005 #4

    LeonhardEuler

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    Have you found "a" now?
     
  6. Sep 14, 2005 #5
    yep a=3radical1 rite?
     
  7. Sep 14, 2005 #6

    LeonhardEuler

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    You mean [itex]a=\frac{1}{3}[/itex] right?
     
  8. Sep 14, 2005 #7
    yes and the function f(x)=5*1/3^x that's what i had got before but when i put down in the answer and it said "wrong", can u help me ??
     
  9. Sep 14, 2005 #8

    LeonhardEuler

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    That seems right. Try 5*(1/3)^x, with the parentheses, maybe.
     
  10. Sep 14, 2005 #9
    beside i still have 1 more problem that is really need help will u able to help me plz?
     
  11. Sep 14, 2005 #10
    yes i did every single posible way but it stills said "wrong" or the answer is not accepted.
     
  12. Sep 14, 2005 #11

    LeonhardEuler

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    Alright, what's the other problem?
     
  13. Sep 14, 2005 #12

    LeonhardEuler

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    What was the exact phrasing of the problem?
     
  14. Sep 14, 2005 #13
    the exact phrasing was what i posted it on at first.
     
  15. Sep 14, 2005 #14
    and the other problem is
    compare the functions f(x)=x^5 and g(x)=5^x by graphing both f and g in several viewing rectangles.
    a) find the x-coordinates of the point intersectionof the two curves accurate to two decimal places. and i did the graphing w/ a result of (1.70,1.70) is that rite?
    and part b) find the value of x greater than 5 for which g(x)=400 f(x)
     
  16. Sep 14, 2005 #15
    yeah i got the first problem rite. thanks
     
  17. Sep 14, 2005 #16

    LeonhardEuler

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    No, that's not right, in fact niether function passes through that point. You could graph the two functions in a different window to find the true intersection, but it is clear by inspection that x=5 will yield 5^5 in both cases. As for the second case, just graph 400x^5 instead of x^5 and see where this intersects g(x).
     
  18. Sep 14, 2005 #17
    i can not find the intersection of 400x^5
     
  19. Sep 14, 2005 #18

    LeonhardEuler

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    Look between x=10 and x=12. y will be between 10^6 and 10^8.
     
  20. Sep 14, 2005 #19
    it's still have no intersection point between 10 or 12-13
     
  21. Sep 14, 2005 #20

    LeonhardEuler

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    The intersection point is between x=10 and x=12. Just make sure you're looking at y values around between 10^6 and 10^8.
     
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