anonymous
  • anonymous
Exponential equation. Points (-2,9) (0,1). My answer y=3(1/3)^x. Am I correct?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Well \[3*(1/3)^{-2} = 3*3^2 = 9*3 = 27 \ne 9\] so nope. That can't be right.
anonymous
  • anonymous
hmm well lets see. First I got a=9/b^2. is that correct?
anonymous
  • anonymous
From the second point we have \[1 = ab^0\] Since \[b^0=1\rightarrow 1 = ab^0 = a*1 \rightarrow a = 1\]

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anonymous
  • anonymous
Try plotting y = 2^x, y = 2^(-x), etc. That may give you ideas about the base and the exponent,
anonymous
  • anonymous
oh well I was taught to use the first point first to find out what a equals for the b equation.
anonymous
  • anonymous
....? plotting?
anonymous
  • anonymous
Anytime you have x = 0 for an exponential equation, the y value for that point is the value of a.
anonymous
  • anonymous
Because any base raised to the 0 power equals 1.
anonymous
  • 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.
anonymous
  • anonymous
Ok. Lets break it down. We have 2 points (-2,9) and (0,1) That gives 2 equations: \[9 = ab^{-2}\] and \[1 = ab^0\]
anonymous
  • anonymous
With me so far?
anonymous
  • anonymous
si.
anonymous
  • anonymous
Ok. So what is \[b^0=?\]
anonymous
  • anonymous
this is where your throwing me off. Don't i have to use the first equation for a first?
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
Nope. Not a trick.
anonymous
  • anonymous
okay well to me its a different way so continue senor.
anonymous
  • anonymous
If you really want we can do it your way.
anonymous
  • anonymous
no I'm open to your idea. your the smarter one here so continue.
anonymous
  • anonymous
Lets do it your way and then we'll do mine and see that we get the same thing.
anonymous
  • anonymous
okay sounds good to me. :0]
anonymous
  • anonymous
Are you familiar with this notation? \[\rightarrow\]
anonymous
  • anonymous
does that mean equals or references to the next step?
anonymous
  • anonymous
not equals. It means 'implies', 'therefore', or 'leads to'
anonymous
  • anonymous
So something like \[ A = 5\] \[ A +B = 10 \rightarrow 5 + B = 10 \rightarrow B = 5\]
anonymous
  • anonymous
Make sense?
anonymous
  • anonymous
okay continue :0]
anonymous
  • anonymous
So using the first equation and solving for a: \[9 = ab^{-2}\] \[\rightarrow 9b^2 = ab^{-2}b^2\] \[\rightarrow9b^2 = ab^0 \rightarrow a = 9b^2\]
anonymous
  • anonymous
Follow that?
anonymous
  • 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)
anonymous
  • anonymous
then solve for b and then plug in b for the original equation.
anonymous
  • anonymous
Ah. No because \[b^{-2} = \frac{1}{b^2}\]
anonymous
  • anonymous
So you need to multiply by b^2 in order to get a by itself, not divide.
anonymous
  • anonymous
oooooh so thats where I missed the mark. Okay i get it so if i have a negative, instead of dividing I multiply.
anonymous
  • 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? \[a^b * a^c = a^{(b+c)}\]
anonymous
  • anonymous
yes and when you divide you subtract.
anonymous
  • anonymous
Right. So here's the million dollar question. What is \[k^0\]
anonymous
  • anonymous
0 right?
anonymous
  • anonymous
Nope. 2^4 = 16 2^3 = 8 2^2 = 4 2^1 = ? 2^0 = ?
anonymous
  • anonymous
shoot I change my mind its 1!! now wheres my money?
anonymous
  • anonymous
heheh.
anonymous
  • anonymous
Right. So if we have \[ab^2 = 9\] We can multiply both sides by b^-2 \[a*b^2*b^{-2} = 9*b^{-2}\]
anonymous
  • 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 \[a = 9b^{-2} \]
anonymous
  • 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
anonymous
  • anonymous
okay gotcha.
anonymous
  • anonymous
So we end up with \[a = 9b^2\]
anonymous
  • anonymous
And we plug that into the second equation.
anonymous
  • anonymous
So what do you get when you plug in 9b^2 for a in the second equation.
anonymous
  • anonymous
i got 1/9=b^2
anonymous
  • anonymous
after dividing 9 from 9b^2 of course.
anonymous
  • anonymous
So b is?
anonymous
  • anonymous
b=1/3 Like I had before...
anonymous
  • anonymous
Yeah. So now plug that in for b in either of the two equations.
anonymous
  • anonymous
then i got a=1. and a full answer of y=1(1/3)^x
anonymous
  • anonymous
Yep.
anonymous
  • 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?
anonymous
  • anonymous
what do you mean? I got the whole equation right didnt i?
anonymous
  • anonymous
Yes. I'm trying to show my original idea.
anonymous
  • 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?
anonymous
  • 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?
anonymous
  • anonymous
no just as long as the answer is correct. :0]
anonymous
  • 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.
anonymous
  • anonymous
oh okay yeah I get it.
anonymous
  • anonymous
Because \[ab^0= k\rightarrow a*1 = k \rightarrow a = k\]
anonymous
  • anonymous
Which means you can jump right to finding b and save some time.
anonymous
  • 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!!
anonymous
  • anonymous
Anytime =)

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