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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?
    (19 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!
      (25 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?
    (7 votes)
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  • 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.
    (6 votes)
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    • stelly blue style avatar for user Kim Seidel
      You must use parentheses. If you didn't type in the parentheses as: b^(2/3), then you will get (b^2)/3. Most calcuators (and this website) assume the exponent is one integer unless you use the parentheses to include the entire fraction.
      (7 votes)
  • duskpin sapling style avatar for user Kuai Liang
    how do you type the answer in the practice?
    (5 votes)
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  • male robot hal style avatar for user caldwelljt
    !#$!#@$#@$

    so yea, i made myself do it before i watched... it took me like an hour, and then felt like banging my head on the wall for missing the whole x^(a-b)

    anywho, i went about it a bit diff...
    1. multiplied both sides by m^(!/3)
    2. adjusted the term on the right to m^(3/9)
    3. Ended up with m^(7/9) = m^((k+3)/9)
    4. did log_m (something i learned later), on both sides to remove m
    result: 7/9 = (k+3)/9
    5: multiplied both sides by 9:
    result: 7 = k + 3
    6. subtracted 3 from both sides:
    7 - 3 = k + 3 - 3
    result: 4 = k

    anyways, i'm glad i figured it out, and i felt it was worth the effort to do it on my own, but i really want to know how to recall things I don't always remember, like how x^a/x^b = x^(a-b)

    Is it just drills? if so, where can i do a whole bunch with instant feedback like i get here on the quizzes?
    (4 votes)
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  • blobby green style avatar for user jeremy.a.gant
    I understand how Sal gets his answer, but I tried it a different way and got it wrong. Where am I going wrong?
    m^(7/9)/m^(1/3) = m^(k/9)
    multiply both sides by m^(1/3):
    m^(7/9) = m^(1/3) x m^(k/9)
    m^(7/9) = m^(3/9 + k/9)
    m^(7/9) = m^(3k/9)
    7/9 = 3k/9
    7 = 3k
    k = 7/3
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      The error is in adding the fractions: m^(3/9 + k/9). The result becomes m^[(k+3)/9], not m^(3k/9). You multiplied the numerators instead of adding them.

      It's also a lot easier if you just take the given problem: m^(7/9)/m^(1/3) and since it is division, subtract the exponents.
      m^(7/9)/m^(1/3) = m^(7/9-1/3) = m^(7/9-3/9) = m^(4/9)

      Hope this helps.
      (5 votes)
  • leaf green style avatar for user Neil Lactawan Caneda
    what properties does make m^4/9=m^k/9 to 4/9=k/9 ?
    (3 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
      (5 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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  • winston default style avatar for user michael..chen
    I don't know if I'm getting this wrong, but this means that when you have a fraction in the form of a^x / a^y, you can rewrite it as a^(x-y) ?
    (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.