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Simplifying Problem

  1. Jan 9, 2016 #1
    In the attached picture, how is it justified to factor out an 'i' in the exponent of the 2 and the 5?
     

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  2. jcsd
  3. Jan 9, 2016 #2

    fresh_42

    Staff: Mentor

    Do you know what ##2^i## means?
     
  4. Jan 9, 2016 #3
    Apparently I do not. To clarify, I'm not working with imaginary, "i" is just the variable they choose.
     
  5. Jan 9, 2016 #4

    fresh_42

    Staff: Mentor

    I know. And with another letter: ##2^n## is simply a short form of multiplying ##2## with itself ##n## times.
    ##2^0 = 1## (convention), ##2^1 = 2, 2^2 = 2 \cdot 2 = 4, 2^3= 2 \cdot 2 \cdot 2 =8## and so on.
     
  6. Jan 9, 2016 #5
    Okay. But I'm still not sure on why they can factor out ##\frac{2^{i}}{5^{i}}## algebraically. I get that you can take the 2 out from the top and the 5 from the bottom but how do they get an exponent too?
     
  7. Jan 9, 2016 #6

    fresh_42

    Staff: Mentor

    That's why ##(\frac{a}{b})^n = \frac{a}{b} \cdot ... \cdot \frac{a}{b} (n ## times ##) = \frac{a \cdot ... \cdot a }{b \cdot ... \cdot b} ## each ## n ## times.
    ##(\frac{2}{5})^3 = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{2 \cdot 2 \cdot 2}{5 \cdot 5 \cdot 5} = \frac{2^3}{5^3}##
    And ##2## is contained in every single factor of the nominator and ##5## in every single factor of the denominator.
     
  8. Jan 9, 2016 #7
    Sorry, I still don't follow.
     
  9. Jan 9, 2016 #8

    fresh_42

    Staff: Mentor

    ##2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2i = (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot \cdot \cdot (2 \cdot i) = [ 2 \cdot 2 \cdot 2 \cdot \cdot \cdot (i ## times ##) \cdot \cdot \cdot 2] \cdot [1 \cdot 2 \cdot 3 \cdot \cdot \cdot i] = 2^i \cdot i!##
    and the same with ##5## in the denominator. Then ##i!## cancels out and ##\frac{2^i}{5^i} = (\frac{2}{5})^i## is left.
     
  10. Jan 9, 2016 #9
    I think I understand now! I feel pretty stupid lol. Much appreciated.
     
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