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Rewriting quotient of powers (rational exponents)

Sal rewrites the expression m^(7/9) / m^(1/3) as a single exponential term m^(4/9).

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  • leaf blue style avatar for user Hennessey Venom F5
    This video goes to fast. Can anyone explain to me how all the steps Sal did with X actually work?
    (13 votes)
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    • leafers tree style avatar for user ACgolfgirl
      Lamborghini Huracan, I thought it was somewhat confusing myself!

      Here it is...

      So the point of that messy work was to show the relationship between dividing variables with exponents. If the base (here it is x) is the same in both above and below the fraction, you can subtract the bottom exponent from the top exponent and stick it next to x for the answer. ( Example: x ^7 / x ^4 = x ^3 <- which is 7-4 ) Same thing when he shows x ^a / x ^b. Since the base of x is constant, it simplifies to x ^a-b.

      Sal then goes on to the situation of the fraction itself. x ^a / x^b is the same thing as x^a multiplied by 1/x^b. Right? Then 1/x^b can be simplified to x^-b. The negative exponent represents that it is put under 1. ( Example: a^-4 = 1/a^4 )

      So since it is now been replaced with x^-b, it's now x^a multiplied by x^-b.

      Now with multiplying variables with exponents, the rule is similar. If the bases are the same, you can add the exponents. Since the base of x is constant, you can add "a" and "-b", which is x^a-b.

      This just shows you the background proof for the exponent rule of dividing x^a by x^b.

      Hope this isn't too confusing and that it helps! If you know the exponent rule for dividing numbers with exponents that's all you need to remember, not the background proof!
      (18 votes)
  • leafers seedling style avatar for user Andreas Henrik Andersson
    At Sal writes m^7/9+^1/3=m^k/9. I am confussed on how Sal got rid of the m under the division line. why isn't it m^7/9*m^1/3=m^k/9?
    (6 votes)
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  • orange juice squid orange style avatar for user Cierra
    How would you solve 1 over z to the -1/2 power?
    (2 votes)
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    • aqualine ultimate style avatar for user AD Baker
      Cierra,

      I'm not sure I understand what you are asking, but I will try to answer. I believe you are asking how to solve

      1/(z^(-1/2))

      The negative sign in the exponent indicates that you should take the inverse (move the term to the numerator) and drop the negative sign. Like so:

      (z^(1/2))/1

      Then, simplify

      z^1/2
      (6 votes)
  • aqualine seedling style avatar for user Rassul Kuatov
    Hello everyone.
    Could you please explain to me how to type the fraction exponents in the practice session after this video?
    For example, I am trying to type b^(2/3) but always end up (b^2)/3. Is there something wrong with the program?
    Thank you very much.
    (2 votes)
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  • leaf green style avatar for user Neil Lactawan Caneda
    what properties does make m^4/9=m^k/9 to 4/9=k/9 ?
    (2 votes)
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    • mr pink green style avatar for user David Severin
      I am not sure there is a property that covers it, but it is logical and can be easily proven. Divide by m^(k/9) to get m^(4/9)/m^(k/9) = 1, division with same base means subtract exponents, so m^(4/9 - k/9) = 1. Anything to the 0 power is 1, so 4/9 - k/9 = 0 , thus 4/9 = k/9.
      (2 votes)
  • duskpin sapling style avatar for user Chelsea B.
    In the practice session after this video, I had this problem to solve.
    b^4 * b^1/4=? I added the powers of 4 and 1/4 to get 17/4, because 4/1=16/4 I added to get 17/4. I ended up getting that wrong, so I looked at the hints and its 15/4 why?
    (1 vote)
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  • duskpin sapling style avatar for user Kuai Liang
    how do you type the answer in the practice?
    (2 votes)
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  • female robot grace style avatar for user Juleysi  Rosario
    why are all the formulas a+b or a-b?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      Exponents are shortcuts for multiplication, so x^3 = x*x*x. If you multiply x^3*x^2, you have x*x*x * x*x = x^5, thus the exponents were added 3 + 2 = 5.
      If you are dividing x^5/x^2, you have (x*x*x*x*x)/(x*x), and x/x = 1 because anything except 0 divided by itself is 1. Two of the xs cancel out, so you have x*x*x left = x^3. Subtracting 5-2 = 3. So exponents add when you multiply same bases and subtract when you divide same bases.
      (2 votes)
  • female robot ada style avatar for user A.HAM4Life
    why does Sal show us all of the stuff about x^a/x^b=x^a*1/x^-b=x^a-b?
    there's not really a point to that, is there?
    (1 vote)
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  • spunky sam blue style avatar for user Mang Kim
    How would you solve 1/z to the power of -1/2
    (1 vote)
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Video transcript

- [Voiceover] So we have an interesting equation here. Let's see if we can solve for k, and we're going to assume that m is greater than zero. Like always, pause the video. Try it out on your own, and then I will do it with you. All right, let's work on this a little bit. So you can imagine that the key to this is to simplify it using our knowledge of exponent properties, and there's a couple of ways to think about it. First, we can look at this rational expression here, m to the 7/9 power divided by m to the 1/3 power. And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. And actually comes straight out of the notion that x to the a over x to the b, x to the a over x to the b, is the same thing as x to the a times one over x to the b, which is the same thing as x to the a times... One over x to the b, that's the same thing as x to the negative b, which is going to be the same thing as... If I have a base to one exponent times the same base to another exponent, that's the same thing as that base to the sum of the exponents, a plus negative b which is just gonna be a minus b. So, we got to the same place. So, we can re-write this as... So, we can re-write this part as being equal to m to the 7/9 power minus 1/3 power is equal to, is equal to m to the k over nine. And I think you see where this is going. What is 7/9 minus 1/3? Well, 1/3 is the same thing, if we want to have a common denominator, 1/3 is the same thing as 3/9. So, I can re-write this as 3/9. So 7/9 minus 3/9 is going to be 4/9. So, this is the same thing as m to the... M to the 4/9 power is going to be equal to m to the k-ninths power. So, 4/9 must be the same thing as k-ninths. So, we can say 4/9 is equal to k-ninths. Four over nine is equal to k over nine, which tells us that k must be equal to four, and we're all done.