anonymous
  • anonymous
Hi! I have to put down this formula y = ab^x and I need help with it.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Yes you are. Next, I would divide the second equation by the first equation.
anonymous
  • anonymous
so 4 by 16 or the whole things?
anonymous
  • anonymous
IN order to maintain equality you have to divide both sides. Like this:\[\frac{ 16 }{ 4 } = \frac{ ab^{-3} }{ ab^{-1} }\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so b = 4 then
anonymous
  • anonymous
Well, carrying out the division above, you'll get\[4 = b^{-2} = \frac{1}{b^2}\]Can you solve for b?
anonymous
  • anonymous
um sorry I don't know what exactly to do... i'm trying to think
anonymous
  • anonymous
Let's back up a bit and make sure you're OK with the division.
anonymous
  • anonymous
1/2
anonymous
  • 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
  • anonymous
You are correct, b = 1/2
anonymous
  • anonymous
oh okay yeah I see duh lol
anonymous
  • anonymous
You're doing a good job. Now, substitute b=1/2 into either of the given equations and solve for a.
anonymous
  • anonymous
The first one might be easier to work with\[4=a \left( \frac{ 1 }{ 2 } \right)^{-1}\]
anonymous
  • anonymous
okay um 2?
anonymous
  • anonymous
Correct. Well done.
anonymous
  • anonymous
how would the second equation be set up? same thing just replace the 4?
anonymous
  • 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
  • anonymous
or don't we have to do 16/4?
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
okay so what's next?
anonymous
  • 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
  • anonymous
You have a and b. Put them into the general function\[y=ab^x\]What do you get?
anonymous
  • anonymous
y= 2(1/2)^x ?
anonymous
  • anonymous
Excellent. Well done.
anonymous
  • anonymous
or y= 2(0.5)^x
anonymous
  • anonymous
Whichever you teacher (or courseware) prefers.
anonymous
  • anonymous
I think he wants decimal. Anyways merci beaucoup! I appreciate this so much. You rock
anonymous
  • anonymous
You're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.