Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

....? plotting?

Answer 7

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

Answer 8

Because any base raised to the 0 power equals 1.

Answer 9

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.

Answer 10

Ok. Lets break it down. We have 2 points (-2,9) and (0,1)

That gives 2 equations:
\[9 = ab^{-2}\]
and
\[1 = ab^0\]

Answer 11

With me so far?

Answer 12

si.

Answer 13

Ok. So what is
\[b^0=?\]

Answer 14

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

Answer 15

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.

Answer 16

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.

Answer 17

Nope. Not a trick.

Answer 18

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

Answer 19

If you really want we can do it your way.

Answer 20

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

Answer 21

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

Answer 22

okay sounds good to me. :0]

Answer 23

Are you familiar with this notation?
\[\rightarrow\]

Answer 24

does that mean equals or references to the next step?

Answer 25

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

Answer 26

So something like
\[ A = 5\]
\[ A +B = 10 \rightarrow 5 + B = 10 \rightarrow B = 5\]

Answer 27

Make sense?

Answer 28

okay continue :0]

Answer 29

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

Answer 30

Follow that?

Answer 31

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)

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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?

\[a^b * a^c = a^{(b+c)}\]

Answer 37

yes and when you divide you subtract.

Answer 38

Right. So here's the million dollar question. What is
\[k^0\]

Answer 39

0 right?

Answer 40

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

Answer 41

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

Answer 42

heheh.

Answer 43

Right. So if we have

\[ab^2 = 9\]
We can multiply both sides by b^-2
\[a*b^2*b^{-2} = 9*b^{-2}\]

Answer 44

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
\[a = 9b^{-2} \]

Answer 45

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

Answer 46

okay gotcha.

Answer 47

So we end up with
\[a = 9b^2\]

Answer 48

And we plug that into the second equation.

Answer 49

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

Answer 50

i got 1/9=b^2

Answer 51

after dividing 9 from 9b^2 of course.

Answer 52

So b is?

Answer 53

b=1/3 Like I had before...

Answer 54

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

Answer 55

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

Answer 56

Yep.

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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?

Answer 61

\[1=ab^0\] 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?

Answer 62

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

Answer 63

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.

Answer 64

oh okay yeah I get it.

Answer 65

Because
\[ab^0= k\rightarrow a*1 = k \rightarrow a = k\]

Answer 66

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

Answer 67

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!!

Answer 68

Anytime =)