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### Course: Algebra 2>Unit 6

Lesson 2: Properties of exponents (rational exponents)

# Rewriting quotient of powers (rational exponents)

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

## Want to join the conversation?

• This video goes to fast. Can anyone explain to me how all the steps Sal did with X actually work?
• 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!
• 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?
• Sal is using the property of exponents for division. When we divide and have a common base, we subtract the exponents: m^7 / m^2 = m^(7-2) = m^5

Sal's problem is a little more complicated because the exponents are fractions. But, he is using the same property: m^(7/9) / m^(1/3) = m^(7/9-1/3)

In your version: m^7/9*m^1/3, you have changed the division into multiplication which can't be done without changing the exponent on the 2nd m to be -1/3.

Hope this helps.
• 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.
• 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.
• how do you type the answer in the practice?
• You can use the different functions in the answer box.
• !#\$!#@\$#@\$

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?
• I would recommend doing a lot of practice problems!
(1 vote)
• what properties does make m^4/9=m^k/9 to 4/9=k/9 ?
• 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.
• How would you solve 1 over z to the -1/2 power?
• 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
• 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
• 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.