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znimon Group Title

If f(x) = 8^x, show that (f(x+h) - f(x))/h = 8^x((8^h-1)/h) I am not sure of the properties at play here. I have the answer. If someone would refer some khan academy videos for further explanation that would be appreciated.

  • one year ago
  • one year ago

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  1. dmgrass Group Title
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    Just laws of exponents: \[(f(x+h)-f(x))/h=(8^{x+h}-8^x)/h=(8^x 8^h-8^x)/h=8^x(8^h-1)/h\]

    • one year ago
  2. znimon Group Title
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    As i don't know the names of those laws I can't look up videos on them. Will you give me the names?

    • one year ago
  3. dmgrass Group Title
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    I always refer to them as the "laws of exponents" ... you can search for this quoted phrase on the web or YouTube.

    • one year ago
  4. zepdrix Group Title
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    So the funky looking thing they gave you is called the "Difference Quotient" Remember back to algebra using the uhhh, ugh i forget what it's called. To find the slope of a line.\[m=\frac{ y_2-y_1 }{ x_2-x_1 }\] It's the same thing, it's the slope of a line, given 2 points. But now we're using function notation so it's a little bit fancier :) The distance between x_2 and x_1, we call that h. So now our y (which is now f(x)), the second y value will be the FIRST y value + the distance h that we traveled. f(x+h) - f(x).

    • one year ago
  5. zepdrix Group Title
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    |dw:1350357859691:dw| So this is what your initial setup should look like when you get everything plugged in :) Make sense? :o

    • one year ago
  6. znimon Group Title
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    I don't understand what property/properties of exponents allow me to conclude 8^(x + h) = 8^x * 8^h

    • one year ago
  7. znimon Group Title
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    Yeah

    • one year ago
  8. zepdrix Group Title
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    Hmm it's one of the exponent laws. The product of two terms with the same BASE, can be written as the sum of their exponents. I Dunno what the specific law is called :D It's a good one to remember though. \[a^b*a^c=a^{b+c}\]

    • one year ago
  9. zepdrix Group Title
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    We're applying this rule in reverse in this case.

    • one year ago
  10. zepdrix Group Title
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    k

    • one year ago
  11. znimon Group Title
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    That cleared things up, thanks.

    • one year ago
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