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anonymous

  • one year ago

Given the function f(x) = log2(x + 6), find the value of f^−1(3).

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  1. anonymous
    • one year ago
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    I really need help with understanding this.

  2. anonymous
    • one year ago
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    Answer Choices: a. f−1(3) = 2 b. f−1(3) = 3 c. f−1(3) = 9 d. f−1(3) = 18

  3. anonymous
    • one year ago
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    @zepdrix help please

  4. anonymous
    • one year ago
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    @SkaterBoyShawn can you help me?

  5. anonymous
    • one year ago
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    yes

  6. anonymous
    • one year ago
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    okay good

  7. freckles
    • one year ago
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    \[f^{-1}(3)=a \implies f(a)=3 \\ \text{ so you need to solve } \log_2(a+6)=3 \text{ for } a \]

  8. anonymous
    • one year ago
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    @freckles in my notes it was saying how i had to find the inverse function or something like that

  9. anonymous
    • one year ago
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    let me solve it 1sec

  10. anonymous
    • one year ago
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    ok @SkaterBoyShawn

  11. freckles
    • one year ago
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    You can find the inverse function then plug in 3 or you can just solve the equation above for a

  12. anonymous
    • one year ago
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    ugh im still confused

  13. anonymous
    • one year ago
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    i think i got the answer

  14. freckles
    • one year ago
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    on what part?

  15. freckles
    • one year ago
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    like why \[f^{-1}(3)=a \implies f(a)=3 ? \\ \text{ or on solving } \log_2(a+6)=3 \text{ for } a?\]

  16. anonymous
    • one year ago
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    how? @SkaterBoyShawn and @freckles everything like what do you do with log2?

  17. freckles
    • one year ago
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    you want to write in the equivalent exponential form

  18. freckles
    • one year ago
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    recall \[\log_b(x)=y \implies b^{y}=x\]

  19. anonymous
    • one year ago
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    i dont get any of this at all. Math is my weakest subject

  20. freckles
    • one year ago
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    so you don't know how to compare \[\log_b(x)=y \text{ to } \log_2(a+6)=3 \\ \text{ then use that } \log_b(x)=y \implies b^y=x \\ \text{ to write } \log_2(a+6)=3 \text{ in exponential form }\]?

  21. freckles
    • one year ago
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    |dw:1433709916685:dw| what is in place of the b? in place of the x? in place of the y?

  22. anonymous
    • one year ago
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    i dont know how to write that in exponential form.

  23. anonymous
    • one year ago
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    f⁻¹(3) = 2³-6 = 2 if u in verse it it i u would get 2^y = x+6 x = 2^y-6 f⁻¹(x) = 2ˣ-6

  24. freckles
    • one year ago
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    I'm just asking you to identify b,x, and y in that comparison

  25. freckles
    • one year ago
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    |dw:1433710114097:dw| do you not see that b is 2? can you try to identify what is place of x and y now?

  26. anonymous
    • one year ago
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    i see that now. x is 6 and y is 3.

  27. freckles
    • one year ago
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    well x is everything in that log thing so x is a+6 not just 6

  28. anonymous
    • one year ago
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    oh okay

  29. freckles
    • one year ago
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    \[\log_b(x)=y \implies b^y=x \\ \text{ so we can use this to write } \log_2(a+6)=3 \implies 2^{3}=a+6\]

  30. freckles
    • one year ago
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    can you solve 2^3=a+6 for a ?

  31. anonymous
    • one year ago
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    a=2

  32. freckles
    • one year ago
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    yep

  33. anonymous
    • one year ago
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    y would be f⁻¹(x) = 2ˣ-6 not sure yet for b is

  34. freckles
    • one year ago
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    so since f(2)=3 then f^(-1)(3)=2

  35. anonymous
    • one year ago
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    oh okay then

  36. freckles
    • one year ago
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    And I think @SkaterBoyShawn is trying to attempt to find the inverse function and then plug in there instead which is cool too

  37. anonymous
    • one year ago
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    so for problems like that all you have to do is plug in the numbers and solve for a?

  38. freckles
    • one year ago
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    \[y=\log2(x+6) \\ 2^y=x+6 \\ x=2^y-6 \\ f^{-1}(x)=2^x-6\] his inverse function is right

  39. anonymous
    • one year ago
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    correct @freckles

  40. freckles
    • one year ago
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    so if you have done it this way you can just replace the x with 3

  41. anonymous
    • one year ago
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    thank you

  42. anonymous
    • one year ago
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    do u guys understand the the inverse function

  43. anonymous
    • one year ago
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    no

  44. anonymous
    • one year ago
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    no u don't

  45. freckles
    • one year ago
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    yeah the function f(x)=log_b(x) is one to one so it does have an inverse function to find something like this: \[f^{-1}(3) \] where f^(-1) is the inverse of f you could say well I know the following \[f^{-1}(3)=a \implies f(a)=3\] so yes in the original you can replace x with a (or just leave the x there if you are feeling lazy) and then solve for a (or x if you left a x instead)

  46. anonymous
    • one year ago
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    @freckles u understand

  47. freckles
    • one year ago
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    understand what?

  48. anonymous
    • one year ago
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    oh okay I got it now i think @freckles

  49. freckles
    • one year ago
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    In general: Let's say you have the following problem: \[\text{ Find } f^{-1}(m) \text{ if } f(x)=\log_b(x) \\ \\ \text{ assume } f^{-1}(m)=a \text{ well this means } f(a)=m \\ \\ \text{ so you replace } x \text{ with } a \text{ like so } \\ f(a)=\log_b(a) \\ \text{ now remember } f(a)=m \\ \text{ so we can replace } f(a) \text{ with } m \\ m=\log_b(a) \\ \text{ and solve for } a \\ b^m=b^{\log_b(a)} \\ b^m=a \\ a=b^m \\ \text{ so remember } a=f^{-1}(m) \\ \text{ and that is what we were aiming to find } \\ f^{-1}(m)=a=b^m \\ f^{-1}(m)=b^m\]

  50. freckles
    • one year ago
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    notice the is the inverse function of f(x) evaluated at x=m

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