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I need to find the function of f with given data

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi

    I need to determine ragulatory for f, where the expiration time is measured as the function of the temperature.

    The function type is f(t)=b*at

    i get two pair of datas which are that for -15 °C the expiration day will be after 200 and by -5 ° the expiration day will be after 10 days

    2. Relevant equations
    I know that t is the temperature and f(t) must be the days, and i need to determine a regulatory for f, which means i need to find a and b i think

    so i should maybe put my given informations in f(t)=b*at which will be

    200=b*a-15

    and

    10=b*a-5

    then i shall isolate b

    200/a-15 =b and 10/a-5 =b

    these to get put up agains each other so a can be solved.

    200*a-5 =10*a-15

    now i solve a

    200=10*a-20

    a=tenth√(200/10)

    a=1.16

    But it is wrong and i dont get why so can someone please help me??


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 7, 2013 #2

    gneill

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    Staff: Mentor

    Oops! Check your exponent work; you want to multiply both sides by a5.

    Otherwise, you seem to be doing okay :smile:
     
  4. Jan 7, 2013 #3


    thank youu but sorry i cannot see on which step of the ones i have to do it but is it right if i think what you are telling me is what happens in the third last of my steps?

    200*a-5*a5 =10*a-15*a5?? but i dont know how to solve it when it is written like this
     
  5. Jan 7, 2013 #4

    gneill

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    Staff: Mentor

    Yes, that's where I see a problem. You wanted to get rid of the a-5 on the left so you divided both sides by a-5, which is the same thing as multiplying both sides by a5. But you made a mistake handling the exponents on the right hand side.

    What's ##a^{-15}\cdot a^5## ?
     
  6. Jan 7, 2013 #5
    that would be a-75?

    so then i solve a

    200=10*a-75

    a=-75th √(200/10)

    but is there not a rule saying that i cannot take the -xsquaroot of something and i have -75?



    and should i also keep going and find b like or does one thing need to be undefined to have a function?
     
  7. Jan 7, 2013 #6

    gneill

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    Staff: Mentor

    Okay, you should review your laws of exponents:

    ##a^x \cdot a^y = a^{x+y}##

    ##\frac{a^x}{a^y} = a^{x-y}##
     
  8. Jan 7, 2013 #7

    then i say -15 + 5 which would be -10? but you know its the - infront of 10 thats bugging me...because it cant be a minus since i have to take the square root of it later
     
  9. Jan 7, 2013 #8

    gneill

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    No, you shouldn't have to take the square root of a negative number. From your previous work, and repairing the exponent math, the equation is:

    ##200 = 10 a^{-10}##

    ##20 = a^{-10}##

    You can proceed using exponent rules, or use logarithms.
     
  10. Jan 7, 2013 #9
    well im not sure about how to use the logarithm, but i tried solving it with a program that we also use, and what do i do if i get 5 relsults put in a parensesis and under each other

    0
    0
    1,075
    -0,538+0,93li (i dont know what li strands for)
    -0,538-0,93li

    i believe i should use that a is 1,075 so do i need to find the b too? because i dont get how to make an f function out of it now
     
  11. Jan 7, 2013 #10

    gneill

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    The "i" stands for ##\sqrt{-1}##. Those are called "imaginary" numbers (as opposed to the "real" numbers that you're used to), and you'll learn about them at a later date.

    What you're looking for here will be real numbers, so when you take the nth root of a real number you want what's called the "principle value".

    Here you can continue to simplify your expression to make it simpler to solve.

    ##20 = a^{-10}##

    ##20^{-\frac{1}{10}} = a~~~~~~## {raised both sides to the -1/10 to isolate a}

    ##a = \frac{1}{20^{1/10}}~~~~~~## {moved "20" to denominator to clear negative}

    So you need to find the 10th root of 20, then take its inverse.

    Once you have a value for a you can plug it into either of the original expressions to solve for b.
     
  12. Jan 8, 2013 #11
    i dont get how you go from a-10 to -1/10 ? how do you get the 1 ?


    ok so i got the 10th root of 20 to be 1,349


    and i put it in the expression for b so


    200/1,349-15=1,783*104

    and

    10/1,349-5=44,675

    but this shows i did something wrong since my two attempts at finding b were not the same or the two pairs of numbers ive got
     
  13. Jan 8, 2013 #12

    gneill

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    Exponent rules (you should become very familiar with them):
    $$(a^x)^y = a^{x\cdot y}$$
    Here the exponent of a is -10 and you'd like it to be 1. So ##(-10) \cdot \left(-\frac{1}{10}\right) = 1##. The same operation is applied to the 20 on the other side of the equation in order to keep the equation balanced.
    Hmm, that seems unlikely. Surely 20th root of 20 should be less than 20 itself?
    How did you obtain the value?
     
  14. Jan 8, 2013 #13

    I said the 10th root of 200/10...but i know its wrong now...i just tried again and got a = 0,741 isnt that right and b=2,3?

    I know i might be irritating because i keep asking and i really dont hope im bothering you too much but i really still dont get the exponent rule you used last, they are even shown in my math book but i simply dont get it.

    yes i would like it to be one but how did you know that i would have to say - 1/10 ? i mean i wouldnt know that you had to use a 1, and when i compare what is done it does not look like that rule at all ax*y i mean it even contains a division where we are using multiplikation
     
  15. Jan 8, 2013 #14

    gneill

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    Yes, that looks good.
    Okay, let's consider the object ##a^{-10}##. It's exponent is currently -10. If I wish to make the exponent 1, I need to divide it by -10, right? But dividing by -10 is the same as multiplying by ##\frac{1}{-10}##. Putting it in the form of the given exponent rule, letting ##x = -10## and ##y = -\frac{1}{10}##, then

    $$(a^x)^y = a^{xy} = a^{-10 \frac{1}{-10}} = a^\frac{-10}{-10} = a^1 = a$$

    So I chose -1/10 in order to clear the -10 from the exponent of a.

    And of course, if you do something to one side of an equation, you have to do the same thing to the other side in order to retain equality.
     
  16. Jan 8, 2013 #15
    so if i divided -10 by -10 it would also be the right and same result?
     
  17. Jan 8, 2013 #16

    gneill

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    Staff: Mentor

    Sure. Note that on the other side of the equation you have 20, with an assumed exponent of 1. Dividing the exponent by -10 gives ##20^{-1/10}##, the same as before.
     
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