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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    Yes you are. Next, I would divide the second equation by the first equation.

  2. anonymous
    • one year ago
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    so 4 by 16 or the whole things?

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

  4. anonymous
    • one year ago
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    so b = 4 then

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

  6. anonymous
    • one year ago
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    um sorry I don't know what exactly to do... i'm trying to think

  7. anonymous
    • one year ago
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    Let's back up a bit and make sure you're OK with the division.

  8. anonymous
    • one year ago
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    1/2

  9. anonymous
    • one year ago
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    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
    • one year ago
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    You are correct, b = 1/2

  11. anonymous
    • one year ago
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    oh okay yeah I see duh lol

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

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

  14. anonymous
    • one year ago
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    okay um 2?

  15. anonymous
    • one year ago
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    Correct. Well done.

  16. anonymous
    • one year ago
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    how would the second equation be set up? same thing just replace the 4?

  17. anonymous
    • one year ago
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    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
    • one year ago
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    or don't we have to do 16/4?

  19. anonymous
    • one year ago
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    oh okay

  20. anonymous
    • one year ago
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    okay so what's next?

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

  23. anonymous
    • one year ago
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    y= 2(1/2)^x ?

  24. anonymous
    • one year ago
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    Excellent. Well done.

  25. anonymous
    • one year ago
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    or y= 2(0.5)^x

  26. anonymous
    • one year ago
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    Whichever you teacher (or courseware) prefers.

  27. anonymous
    • one year ago
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    I think he wants decimal. Anyways merci beaucoup! I appreciate this so much. You rock

  28. anonymous
    • one year ago
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    You're welcome

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