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Dumb Question: Why Do You Add Exponents When Multiplying?

  1. Nov 2, 2015 #1
    When multiplying x2 and x5, we get x7. Quick, dumb question:

    Why isn't it x10 instead (since we're multiplying and not adding)?
     
  2. jcsd
  3. Nov 2, 2015 #2

    jedishrfu

    Staff: Mentor

    Think of the exponent as counting the number of factors so x^2 is x*x and x^5 is x*x*x*x*x and when multiplied together gives you 7 x's or x^7
     
  4. Nov 2, 2015 #3

    Mark44

    Staff: Mentor

    ##x^2## means ##x \cdot x##, and ##x^2## means ##x \cdot x \cdot x \cdot x \cdot x##, so ##x^2 \cdot x^5## means ##x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x##, right? How many factors of x are there in that last expression?
     
  5. Nov 2, 2015 #4
    This might seem a little dumb, but I figured that multiplying would give you 10 still. Using your example above,

    x2 = x*x
    x5 = x*x*x*x*x

    Wouldn't x2 * x5 = x*x; x*x; x*x; x*x; x*x? Basically, you'd have a pair of x*x's five times. Or, you would get:

    x*x*x*x*x; x*x*x*x*x Or, in other words, two groups of x*x*x*x*x.

    It's the same idea with regular multiplication, where if you have 2X5, then it's like

    || || || || || (five groups of two) or |||||; ||||| (two groups of five)

    I just don't quite understand why we're adding and getting 7 (instead of 10), since the operation is to multiply exponents. Do you guys see what I mean?

    If they want you to add the exponents, then why not just say x2 + x5?
     
  6. Nov 2, 2015 #5

    Mark44

    Staff: Mentor

    No.
    What you have written is ##(x^2)^5##, or 5 factors of ##x^2##, making 10 factors of x, or ##x^{10}##.
    With ##(x^2) \cdot (x^5)## you have two factors of x multiplying 5 factors of x, making 7 factors of x, or ##x^7##.
    No, again. The operation is to multiply ##x^2## and ##x^5##, which results in the exponents being added.
    Because the terms in ##x^2 + x^5## are not like terms (i.e., same exponent), they can't be combined.
     
  7. Nov 2, 2015 #6
    Maybe try looking at it with numbers. For example, when ##x=2##, ##x^2\cdot x^5## is ##4\cdot32=128=2^7##. Try it with some more numbers. Now you will see that indeed, as Mark44 said, ##x^2\cdot x^5=(x\cdot x)\cdot(x\cdot x\cdot x\cdot x\cdot x)## and not what you wrote.
     
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