- anonymous

Hi!
I have to put down this formula y = ab^x and I need help with it.

- chestercat

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- anonymous

Yes you are. Next, I would divide the second equation by the first equation.

- anonymous

so 4 by 16 or the whole things?

- anonymous

IN order to maintain equality you have to divide both sides. Like this:\[\frac{ 16 }{ 4 } = \frac{ ab^{-3} }{ ab^{-1} }\]

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- anonymous

so b = 4 then

- anonymous

Well, carrying out the division above, you'll get\[4 = b^{-2} = \frac{1}{b^2}\]Can you solve for b?

- anonymous

um sorry I don't know what exactly to do... i'm trying to think

- anonymous

Let's back up a bit and make sure you're OK with the division.

- anonymous

1/2

- anonymous

You had\[\frac{ 16 }{ 4 } = \frac{ ab^{-3} }{ ab^{-1} }\]The left hand side is easy. On the right hand side, the a's will cancel out, leaving\[4=\frac{ b^{-3} }{ b^{-1} }\]Are you OK with it so far?

- anonymous

You are correct, b = 1/2

- anonymous

oh okay yeah I see duh lol

- anonymous

You're doing a good job. Now, substitute b=1/2 into either of the given equations and solve for a.

- anonymous

The first one might be easier to work with\[4=a \left( \frac{ 1 }{ 2 } \right)^{-1}\]

- anonymous

okay um 2?

- anonymous

Correct. Well done.

- anonymous

how would the second equation be set up? same thing just replace the 4?

- anonymous

Well, you don't really need it. You already have the answer, but if you chose to work with the second equation instead, it would look like\[16 = a \left( \frac{ 1 }{ 2 } \right)^{-3}\]and you get the same answer, a=2.

- anonymous

or don't we have to do 16/4?

- anonymous

oh okay

- anonymous

okay so what's next?

- anonymous

You need one more step: reject the value b =-1/2
Since \(b^2=\dfrac{1}{4}\\b=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)

- anonymous

You have a and b. Put them into the general function\[y=ab^x\]What do you get?

- anonymous

y= 2(1/2)^x ?

- anonymous

Excellent. Well done.

- anonymous

or y= 2(0.5)^x

- anonymous

Whichever you teacher (or courseware) prefers.

- anonymous

I think he wants decimal. Anyways merci beaucoup! I appreciate this so much. You rock

- anonymous

You're welcome

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