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Question about simple algebraic exponential property

  1. Jul 27, 2009 #1
    I haven't been taking math for 3 years so I have a question about the following:

    is 14m6n2 the same as 2mn x 7m5n

    basically asking this because I am not sure whether or not I can factor the terms out like this

    if this information is insufficient I can post the whole problem. Thanks in advance.
     
  2. jcsd
  3. Jul 27, 2009 #2
    It's not the same

    [tex]14^{m^6 n^2} = (2 x 7)^{m^6 n^2} = 2^{m^6 n^2} 7^{m^6 n^2}[/tex]
     
  4. Jul 27, 2009 #3
    ok so then i do not know how i would arrive at the right answer:

    14m6n2 / 2mn

    i know what the right answer is, and i know how they did it. but i cant seem to grasp the reasoning

    PS: the answer is 7m5n

    I was under the impression that you can only subtract the powers through the same base, but apparently i was wrong over here?

    or are numbers different than variables?
     
  5. Jul 27, 2009 #4
    They behave in the same manner, but you know more properties of them. A variable can be any number.

    For example, "songoku" manipulated 14 to read (2*7). You probably already know that [tex](ab)^{n}=a^{n}b^{n}[/tex]. Thus, we can write:

    [tex]\frac{14^{m^{6}n^{2}}}{2^{mn}}[/tex] = [tex]\frac{(2*7)^{m^{6}n^{2}}}{2^{mn}}=\frac{2^{m^{6}n^{2}}*7^{m^{6}n^{2}}}{2^{mn}}}[/tex].

    Is this equal to [tex]7^{m^{5}n}[/tex]?
     
  6. Jul 27, 2009 #5
    i know with the same base i can reduce, but i dont get 7m5n
     
  7. Jul 27, 2009 #6
    maybe you can post the whole question?
     
  8. Jul 28, 2009 #7
    i did:
    its

    14m6n2 / 2mn

    and the answer being 7m5n
     
  9. Jul 28, 2009 #8
    If so, the answer is wrong

    Just check it : let m = 1 and n = 2

    [tex]\frac{14^{m^6 n^2}}{2^{mn}} = \frac{14^4}{2^2} = 9604[/tex]

    [tex]7^{m^5n} = 7^2 = 49[/tex]
     
    Last edited: Jul 28, 2009
  10. Jul 28, 2009 #9
    So it looks like you really have [tex]\frac{14m^6} {2mn}[/tex] and the simplified answer is [tex]\frac{7m^5} {n}[/tex]

    This should be a lot easier for you to do than what you were doing.
     
  11. Jul 29, 2009 #10
    only thing i can guess is that the book is wrong, the number substitution proves this
     
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