## anonymous 5 years ago Exponential equation. Points (-2,9) (0,1). My answer y=3(1/3)^x. Am I correct?

1. anonymous

Well $3*(1/3)^{-2} = 3*3^2 = 9*3 = 27 \ne 9$ so nope. That can't be right.

2. anonymous

hmm well lets see. First I got a=9/b^2. is that correct?

3. anonymous

From the second point we have $1 = ab^0$ Since $b^0=1\rightarrow 1 = ab^0 = a*1 \rightarrow a = 1$

4. anonymous

Try plotting y = 2^x, y = 2^(-x), etc. That may give you ideas about the base and the exponent,

5. anonymous

oh well I was taught to use the first point first to find out what a equals for the b equation.

6. anonymous

....? plotting?

7. anonymous

Anytime you have x = 0 for an exponential equation, the y value for that point is the value of a.

8. anonymous

Because any base raised to the 0 power equals 1.

9. anonymous

still no comprendo senor.....or senora. You don't use the second equation for a.......gawh I wish math was simple for fooligans like me.

10. anonymous

Ok. Lets break it down. We have 2 points (-2,9) and (0,1) That gives 2 equations: $9 = ab^{-2}$ and $1 = ab^0$

11. anonymous

With me so far?

12. anonymous

si.

13. anonymous

Ok. So what is $b^0=?$

14. anonymous

this is where your throwing me off. Don't i have to use the first equation for a first?

15. anonymous

It doesn't matter. You use whichever is easier. In this case the second equation is easier. This will become apparent when you understand what b^0 is.

16. anonymous

okay. I just want to make sure. I have a test tomorrow and if i use your tricks I don't want to get penalized.

17. anonymous

Nope. Not a trick.

18. anonymous

okay well to me its a different way so continue senor.

19. anonymous

If you really want we can do it your way.

20. anonymous

no I'm open to your idea. your the smarter one here so continue.

21. anonymous

Lets do it your way and then we'll do mine and see that we get the same thing.

22. anonymous

okay sounds good to me. :0]

23. anonymous

Are you familiar with this notation? $\rightarrow$

24. anonymous

does that mean equals or references to the next step?

25. anonymous

not equals. It means 'implies', 'therefore', or 'leads to'

26. anonymous

So something like $A = 5$ $A +B = 10 \rightarrow 5 + B = 10 \rightarrow B = 5$

27. anonymous

Make sense?

28. anonymous

okay continue :0]

29. anonymous

So using the first equation and solving for a: $9 = ab^{-2}$ $\rightarrow 9b^2 = ab^{-2}b^2$ $\rightarrow9b^2 = ab^0 \rightarrow a = 9b^2$

30. anonymous

31. anonymous

Okay question time! Don't you divide by b^2? and keep 9/b^2 for the next equations a? for example: 1=(9/b^2)(b^0)

32. anonymous

then solve for b and then plug in b for the original equation.

33. anonymous

Ah. No because $b^{-2} = \frac{1}{b^2}$

34. anonymous

So you need to multiply by b^2 in order to get a by itself, not divide.

35. anonymous

oooooh so thats where I missed the mark. Okay i get it so if i have a negative, instead of dividing I multiply.

36. anonymous

Think about it this way. You want to move the b over to the other side. You know that multiplying powers of the same base you add exponents right? $a^b * a^c = a^{(b+c)}$

37. anonymous

yes and when you divide you subtract.

38. anonymous

Right. So here's the million dollar question. What is $k^0$

39. anonymous

0 right?

40. anonymous

Nope. 2^4 = 16 2^3 = 8 2^2 = 4 2^1 = ? 2^0 = ?

41. anonymous

shoot I change my mind its 1!! now wheres my money?

42. anonymous

heheh.

43. anonymous

Right. So if we have $ab^2 = 9$ We can multiply both sides by b^-2 $a*b^2*b^{-2} = 9*b^{-2}$

44. anonymous

And we end up with $a*b^{2+ (-2)} = 9b^{-2}$ Which simplifies to $a*b^{0} = 9b^{-2}$ Which simplifies to $a*1 = 9b^{-2}$ Which simplifies to $a = 9b^{-2}$

45. anonymous

Now that wasn't the equation we had. We had $ab^{-2} = 9$ So instead of multiplying by b^-2 we need to multiply by b^2

46. anonymous

okay gotcha.

47. anonymous

So we end up with $a = 9b^2$

48. anonymous

And we plug that into the second equation.

49. anonymous

So what do you get when you plug in 9b^2 for a in the second equation.

50. anonymous

i got 1/9=b^2

51. anonymous

after dividing 9 from 9b^2 of course.

52. anonymous

So b is?

53. anonymous

54. anonymous

Yeah. So now plug that in for b in either of the two equations.

55. anonymous

then i got a=1. and a full answer of y=1(1/3)^x

56. anonymous

Yep.

57. anonymous

So now look at what happens when you plug in 1/3 for b in the second equation. Does it matter what b was?

58. anonymous

what do you mean? I got the whole equation right didnt i?

59. anonymous

Yes. I'm trying to show my original idea.

60. anonymous

oh okay. well no it doesn't matter what b was. So yes you were right. hahaa that all you wanted to hear wasnt it?

61. anonymous

$1=ab^0$ is our second equation from the beginning. You plug in 1/3 and you get a = 1. But does it matter what you plug in for b?

62. anonymous

no just as long as the answer is correct. :0]

63. anonymous

Right. So you can see that for exponential equations if you are given a point (0,k). You already know that a must be k.

64. anonymous

oh okay yeah I get it.

65. anonymous

Because $ab^0= k\rightarrow a*1 = k \rightarrow a = k$

66. anonymous

Which means you can jump right to finding b and save some time.

67. anonymous

so if i do it either way i will end up with the right answer. Okay i'm getting your drift here. thanks a bunch! your so patient and I feel more confident it passing my test. SO thanks it really means a lot!!

68. anonymous

Anytime =)