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anonymous

  • 5 years ago

does anyone know how to find the formula for exponential functions that satisfy these given conditions? f(0)=4; F(1)=2 h(0)=100; h(1)=75 p(1)=4; p(2)=6

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  1. amistre64
    • 5 years ago
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    (0,4) (1,2) we know that an exponent is normally at (0,1) so this has been modified at least by +3

  2. amistre64
    • 5 years ago
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    (0,4-3) (1,2-3) falls below the range tho :) we can try using the points to figure it out :) wrap an exponent to fit

  3. amistre64
    • 5 years ago
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    or find the inverse; with logs :)

  4. amistre64
    • 5 years ago
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    f(4) = 0 ; f(2) = 1 logb(4) = 0 ; logb(2) = 1

  5. amistre64
    • 5 years ago
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    b^1 = 2 suggests that b=2...but its modified :) so that cant be right. b^0 = 4 suggests that we are b^x + 3 y = b^x +3 y-3 = b^x 2-3 = b^(1) b = -1 ??

  6. amistre64
    • 5 years ago
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    y = -(1)^x +3 is what I seem to get :)

  7. anonymous
    • 5 years ago
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    An exponential function has the form: \[y=a.b^x\] You just need to substitute using the two points given in each case, and solve for a, b. I will do the first one in the next post.

  8. amistre64
    • 5 years ago
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    yay!! someone smarter than me :)

  9. amistre64
    • 5 years ago
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    im guessing a = 4 ;)

  10. anonymous
    • 5 years ago
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    For the first one, we have the following two equations: \[4=a.b^0 \rightarrow(1)\] and \[2=a.b^1\rightarrow(2)\]

  11. anonymous
    • 5 years ago
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    So clearly, as amistre64 just said, since b^0=1, we have a=4. Then for equation (2): \[2=4b \implies b={2 \over 4}= {1 \over 2}\]

  12. anonymous
    • 5 years ago
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    So we have a=4 and b=1/2, then the exponential function is: \[f(x)=4({1 \over 2})^x\]

  13. anonymous
    • 5 years ago
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    Does that make sense to you? and @amistre64: I am not smarter than you :)

  14. anonymous
    • 5 years ago
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    The answer does... I'm not sure I understand how you got it though

  15. anonymous
    • 5 years ago
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    Which step don't you understand?

  16. anonymous
    • 5 years ago
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    You first should know that the form of an exponential function is: \[f(x)=a.b^x\] where a and b are constants.

  17. anonymous
    • 5 years ago
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    well for the next one, h would also equil to ^0 so 1 so does that make it F(x)=100x 75^1

  18. anonymous
    • 5 years ago
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    So, your goal from here is to find these two constants using the two given points in each case.

  19. anonymous
    • 5 years ago
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    ?

  20. anonymous
    • 5 years ago
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    okay

  21. anonymous
    • 5 years ago
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    The formula makes sence to me, its just the way the points fit into it that confuses me

  22. anonymous
    • 5 years ago
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    I will tell you how to substitute the points. Let's take the point f(0)=4, that says when you plug x=0, you will get value f(x), h(x) (or whatever you call it) to be 4. So the number between the brackets goes for the exponent.

  23. anonymous
    • 5 years ago
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    So for h(0)= 100... H=1 because anything raised to the zero power is one then times 100 which is 100? =/

  24. anonymous
    • 5 years ago
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    Not quite, you should always think of the formula. Now, the formula looks like: \[h(x)=a.b^x\] Try to use the points to find a and b.

  25. anonymous
    • 5 years ago
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    Let me show you how to use the first point to find a, and then you should try finding b. The point is h(0)=100. So, plug x=0, and 100 for h(x).

  26. anonymous
    • 5 years ago
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    100 x 1?

  27. anonymous
    • 5 years ago
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    So, doing what I just said. We have h(x)=ab^x where x=0, and h(x)=100. that gives: \[h(x)=a.b^x \implies 100=a.b^0\] but b^0=1, then \[100=a(1) \implies a=100\]

  28. anonymous
    • 5 years ago
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    We just did about 75% of the problem. Don't forget that you're looking for values for a and b, and then plug them back in the formula. That's all. Can you use the second point to find b?

  29. anonymous
    • 5 years ago
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    would it be 75^1=75?... that seems too simple.

  30. anonymous
    • 5 years ago
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    Don't forget you want to find b, so try to plug in the formula and tell me what b would be.

  31. anonymous
    • 5 years ago
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    Take a breath :)

  32. anonymous
    • 5 years ago
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    Look closely to the formula: \[h(x)=a.b^x\]

  33. anonymous
    • 5 years ago
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    100 x (75)^1?

  34. anonymous
    • 5 years ago
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    We already found a=100, so we can write it as: \[h(x)=100(b)^x\] RIGHT?

  35. anonymous
    • 5 years ago
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    yes

  36. anonymous
    • 5 years ago
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    So now we will use the second point h(1)=75 to find b. We should put x=1 and h(x)=75. You're close by the way :)

  37. anonymous
    • 5 years ago
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    75/100?

  38. anonymous
    • 5 years ago
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    Exactly. \[75=100(b)^1 \implies 75=100b \implies b={75 \over 100}\] But you can simplify this value a little bit of you like :)

  39. anonymous
    • 5 years ago
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    Simplified b=(3/4).

  40. anonymous
    • 5 years ago
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    So, the equation would look like?!

  41. anonymous
    • 5 years ago
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    I hope I didn't complicate things to you. Anyway, all you have to do now is to substitute these to values a=100, b=3/4 into our original function, you get: \[h(x)=100({3 \over 4})^x \]

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