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anonymous

  • 5 years ago

Question says, "Write an exponential function y=ab^x for a graph that includes the given points. (4,8), (6,32) How do I do this?

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  1. anonymous
    • 5 years ago
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    Deal with the problem as a system of 2 equations in two unknowns. The two equations are: \[8=a. {b^4} \rightarrow(1)\] \[32=a.b^6 \rightarrow(2)\] Now, find the two unknowns a and b.

  2. anonymous
    • 5 years ago
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    Does that make sense to you?

  3. anonymous
    • 5 years ago
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    Try solving the system and tell me what you get.

  4. anonymous
    • 5 years ago
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    ok, I'll try.

  5. anonymous
    • 5 years ago
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    Start by substituting a=8/b^4 into the second equation.

  6. anonymous
    • 5 years ago
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    8/b^4 ^b6 ?

  7. anonymous
    • 5 years ago
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    Hmm. From the first equation we have: \[8=a.b^4 \implies a={8 \over b^4} \rightarrow (3)\]

  8. anonymous
    • 5 years ago
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    Now, take this value of a and substitute it in (2), you get: \[32=a. b^6 \implies 32={8 \over b^4}.b^6 \implies 8b^2=32 \implies b^2=4\] So, b is either 2 or -2.

  9. anonymous
    • 5 years ago
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    Are you following so far?

  10. anonymous
    • 5 years ago
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    yes, I'm looking.

  11. anonymous
    • 5 years ago
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    Now just substitute in (3) by the value of x. You will get a=1/2.

  12. anonymous
    • 5 years ago
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    make sure to keep your eyes open O_O

  13. anonymous
    • 5 years ago
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    So, we have, from equation (3), a=8/b^4, and b=2 or -2. Then: \[a={8 \over 2^4}={8 \over 16}=1/2\]

  14. anonymous
    • 5 years ago
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    Your equation should be one of two (since you have two values for b): \[1) y={1 \over 2}(2)^x \implies 2y=2^x\]

  15. anonymous
    • 5 years ago
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    The second equation is the same, but with (-2)^x.

  16. anonymous
    • 5 years ago
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    Thanks for your help. I'm going to need a lot of practice with exploring exponential models.

  17. anonymous
    • 5 years ago
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    Good luck!! :)

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