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anonymous

  • one year ago

Which functions are correct? f(x) = 24x g(x) = 2x Choose exactly two answers that are correct.

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  1. anonymous
    • one year ago
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  2. zepdrix
    • one year ago
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    \[\large\rm f(x)=2^{4x},\qquad\qquad g(x)=2^x\]So then,\[\large\rm (fg)(x)=2^{4x}\cdot 2^x\]As a reminder, here is our relevant exponential rule:\[\large\rm \color{orangered}{a^x\cdot a^y=a^{x+y}}\]It tells us that when we multiply things that have the same base, we add the exponents.

  3. zepdrix
    • one year ago
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    So what do you think? :o

  4. anonymous
    • one year ago
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    sounds difficult

  5. zepdrix
    • one year ago
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    So these are your two options:\[\large\rm 2^{4x}\cdot2^{x}=2^{4x+x}\]\[\large\rm 2^{4x}\cdot2^{x}=2^{4x\cdot x}\] Based on the rule I posted, which one seems correct?

  6. anonymous
    • one year ago
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    the first one i guess

  7. zepdrix
    • one year ago
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    Good. And our division rule tells us to `subtract` the exponents.\[\large\rm \frac{f}{g}=\frac{2^{4x}}{2^x}=2^{4x-x}\]

  8. anonymous
    • one year ago
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    x-x = 0 ?

  9. zepdrix
    • one year ago
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    It's 4x-x. Not x-x.

  10. zepdrix
    • one year ago
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    4apples - an apple

  11. anonymous
    • one year ago
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    but why is it asking for 2 right answers?

  12. zepdrix
    • one year ago
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    Because we chose one correct answer for \(\large\rm (fg)(x)\), multiplication of these functions. We also need to chose one correct answer for \(\large\rm \left(\frac{f}{g}\right)(x)\), division of these functions.

  13. anonymous
    • one year ago
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    what you think about A snd C?

  14. triciaal
    • one year ago
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    agree

  15. zepdrix
    • one year ago
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    yay good job

  16. anonymous
    • one year ago
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    they correct?

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