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

1. anonymous

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

2. anonymous

so 4 by 16 or the whole things?

3. anonymous

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

4. anonymous

so b = 4 then

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

1/2

9. 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?

10. anonymous

You are correct, b = 1/2

11. anonymous

oh okay yeah I see duh lol

12. anonymous

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

13. anonymous

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

14. anonymous

okay um 2?

15. anonymous

Correct. Well done.

16. anonymous

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

17. 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.

18. anonymous

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

19. anonymous

oh okay

20. anonymous

okay so what's next?

21. anonymous

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

22. anonymous

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

23. anonymous

y= 2(1/2)^x ?

24. anonymous

Excellent. Well done.

25. anonymous

or y= 2(0.5)^x

26. anonymous

Whichever you teacher (or courseware) prefers.

27. anonymous

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

28. anonymous

You're welcome