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

1. anonymous

I really need help with understanding this.

2. anonymous

Answer Choices: a. f−1(3) = 2 b. f−1(3) = 3 c. f−1(3) = 9 d. f−1(3) = 18

3. anonymous

4. anonymous

@SkaterBoyShawn can you help me?

5. anonymous

yes

6. anonymous

okay good

7. freckles

$f^{-1}(3)=a \implies f(a)=3 \\ \text{ so you need to solve } \log_2(a+6)=3 \text{ for } a$

8. anonymous

@freckles in my notes it was saying how i had to find the inverse function or something like that

9. anonymous

let me solve it 1sec

10. anonymous

ok @SkaterBoyShawn

11. freckles

You can find the inverse function then plug in 3 or you can just solve the equation above for a

12. anonymous

ugh im still confused

13. anonymous

i think i got the answer

14. freckles

on what part?

15. freckles

like why $f^{-1}(3)=a \implies f(a)=3 ? \\ \text{ or on solving } \log_2(a+6)=3 \text{ for } a?$

16. anonymous

how? @SkaterBoyShawn and @freckles everything like what do you do with log2?

17. freckles

you want to write in the equivalent exponential form

18. freckles

recall $\log_b(x)=y \implies b^{y}=x$

19. anonymous

i dont get any of this at all. Math is my weakest subject

20. freckles

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

|dw:1433709916685:dw| what is in place of the b? in place of the x? in place of the y?

22. anonymous

i dont know how to write that in exponential form.

23. anonymous

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

I'm just asking you to identify b,x, and y in that comparison

25. freckles

|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

i see that now. x is 6 and y is 3.

27. freckles

well x is everything in that log thing so x is a+6 not just 6

28. anonymous

oh okay

29. freckles

$\log_b(x)=y \implies b^y=x \\ \text{ so we can use this to write } \log_2(a+6)=3 \implies 2^{3}=a+6$

30. freckles

can you solve 2^3=a+6 for a ?

31. anonymous

a=2

32. freckles

yep

33. anonymous

y would be f⁻¹(x) = 2ˣ-6 not sure yet for b is

34. freckles

so since f(2)=3 then f^(-1)(3)=2

35. anonymous

oh okay then

36. freckles

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

so for problems like that all you have to do is plug in the numbers and solve for a?

38. freckles

$y=\log2(x+6) \\ 2^y=x+6 \\ x=2^y-6 \\ f^{-1}(x)=2^x-6$ his inverse function is right

39. anonymous

correct @freckles

40. freckles

so if you have done it this way you can just replace the x with 3

41. anonymous

thank you

42. anonymous

do u guys understand the the inverse function

43. anonymous

no

44. anonymous

no u don't

45. freckles

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

@freckles u understand

47. freckles

understand what?

48. anonymous

oh okay I got it now i think @freckles

49. freckles

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

notice the is the inverse function of f(x) evaluated at x=m