I cannot grasp the rules to this problem! Rational exponents! Reddit algebra r/askmath
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- Опубліковано 22 кві 2024
- Let's discuss how to simplify rational exponents 9^(3/4)*3^(7/2)/27^(3/2). This will help you with your algebra class. Best wishes. This problem is from Reddit r/askmath / xb03wtty2h
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I wouldn't have reduced 10/2 to 5 since you have to convert it back to 10/2 to subtract 9/2
I've found that's usually the best way to go on problems like these. Wait until the end, after you get your final result, to reduce or simplify. That way you're not running the risk of having to undo what you just did, as with this problem, and especially with stuff like radicals in the denominator, you often just have simpler numbers to work with.
Exactly what I was thinking. If you know all denominators are the same it’s a waste of a step to reduce
That's exactly what I did, I kept it as 10/2 knowing I could then subtract 10/2 and 9/2 to get 1/2.
I agree but I suspect he did it intentionally to make sure he showed the proper way to add/subtract fractions at some point in the video, so someone learning wouldn't assume you can always just add numerators or get any other wrong ideas.
Yep.. I like working these problems in my head to see if I can get the correct answer without needing to write it out, and immediately regreted simplifying the numerator when I looked at the denominator lol.
This question contain all the exponent laws
not all of them, it doesn't include the law that says (a/b)^n = (a^n)/(b^n)
Tower of exponent, such as a^(b^c) which is different from (a^b)^c, is not included
Negative exponents, which allows you to reciprocate exponents I.e. 1/a = a^-1
@@Koolplayer12 this is just an extrapolation of b^m/b^n = b^(m-n). In this case a^0 = 1 so a^0/a^1 = a^(0-1) = a^-1, so this one was implicitly included.
@@nasrullahhusnan2289 This is an extension on order of operations, and the convention of not including parentheses when following highest order exponents. I wouldn't list it as an exponent law. I know it typically gets lumped in but that's more for memorization than for functional equivalence.
Thank you! This helped get the old neurons firing again. I didn't realize just how much I remembered about exponents. It clicked for me as soon as you explained the first step.
you're a great teacher, keep it up
hes an excellent teacher...
hes an amazing teacher...
hes an incredible teacher...
hes an astonishing teacher...
no chain
Convert 9^(3/4) into 3^(3/2). Now you have halves in all exponents so you can put the square root on the outside. Convert 27^3 into 3^(3x3) = 3^9:
sqrt(3^3 x 3^7 / 3^9) = sqrt(3^10/3^9) = sqrt(3)
For the first time ever I was able to solve this! Yay!
Nice to see you doing lower level algebra tutorial. I watch a lot of your more advanced stuff. You do good work.
Personally I would have just called the base X and set it to √3 because 3^1/2 was the common denominator.
Doing it that way you get:
X^3 * X^7 ÷ X^9 = X^(3+7-9) = X
Makes it a lot easier than dealing with fractions
yeah, but you are advanced. He took that question because it was an excellent teaching moment, with all the details ❤
It's much easier to just combine the addition/subtraction power rule to say the answer is 3^(3/2 + 7/2 - 9/2) = 3^1/2.
It's essentially doing the same thing, but without introducing x as a variable and trying to find a common denominator (which you then need to multiply by) to get your answer.
Really bprp shouldn't have bothered simplifying 10/2 because it keeps the same form as the 9/2
..its easir only because u already know the rules ... this method teached why ...
edit: u can also go directly to ,Ans = √( 3^(3+7-9) ) ... wich is even simpler
...but it req u to know the rules, its imo far better to teach why this is true..
It should also be noted that you have to be careful applying these rules with negative bases because an even root of a negative number is not in the reals. For example (-2)^(1/2) * (-2)^(2) is not equal to (-2)^1 as you might think from applying b^m * b^n = b^(m+n). So to be more accurate b^m and b^n must both be real numbers. Edit: Changed my example (-2)^(1/2) * (-2)^(3/2) is one of the cases where it does work and equals (-2)^(2).
I agree with what you saying, but isnt your example a bad one, as I do get 4 as the answer using google
@@Mattiasthesecond you're right. bad example. Should have used (-2)^(1/2) * (-2)^(2)
@@theeternalsw0rd Applying b^m * b^n = b^(m+n) to your new example means suspecting it to be (-2)^(5/2). (-2)^1 would relevant for ( (-2)^(1/2) )^2.
P.S.
Would you agree that (-2)^(1/2) * (-2)^2 = 4*i*(2)^(1/2) ?
I never tried exponents like 3/2 until a couple of days ago when one appeared in the video of a similar channel.
It turns out I missed out on a lot of maths education that is routine today.
At 4:00, you could have said that (3³)^(3/2) was (3^(3/2))³, and simplify the fraction by 3^(3/2). You would have been left with 3^(7/2) / 3^(6/2), a short step away from sqrt(3).
Very good explanation
(a^x1)^x2=a^(x1*x2)
a^x1*a^x2=a^(x1+x2)
(arithmetic if x is even) xth root of a=a^(1/x)
Important for middle school students these will make them solve the problems a lot easier
My approach was to first put the whole fraction into a square root and simplify the exponents that way
9^(3/4)*3^(7/2)/27^(3/2) = sqrt(9^(3/2)*3^7/27^3)
Then just simplify the insides
great video im going to sub
I am not a student but I really appreciate these videos as a really enjoyable way to keep my basic maths skills fresh.
Sir plz make a video on highest common factor.plz sir
Here's the idea of highest common factor, or greatest common factor GCF. A related term, is least common multiple (LCM).
Given two numbers (call them capital X & capital Y), you are looking for a number A, such that A*x = X and A*y = Y.
Example: X = 70 and Y = 28.
Factors they both have in common:
1. They are both even, so 2 is a common factor
2. They are both multiples of 7, so 7 is a common factor
3. This means, the product of 7 and 2, is also a factor. That is: 14.
Are there any others? Let's find out, by dividing both by 14:
28/14 = 2
70/14 = 5
Since 2 and 5 don't have any factors in common, this means that 14 is the greatest common factor.
14*5 = 70
14*2 = 28
A=14, x = 5 and y = 2
A*x = 70
A*y = 28
An application of greatest common factor, is simplifying fractions. Given the fraction, 28/70, everybody brought 14 to the party. So we can cancel 14 from both top and bottom, and we're left with 2/5.
What class are you in?
Great
A comment to help with the algorithm so the video gets shared more.
Wonder why my calculator gives a number that, written as an expression, equals the 4th root of 9, which is equal to the square root of 3?
I suspect it's because the original expression gives a positive only answer that equals 4th root of 9, while entering that result expression into the calculator gives both a positive and negative square root of 3 as answer?
One explanation is that it might have floating point rounding errors behind the scenes, that cause it to not recognize 4th root of 9 as reducible. Many calculators that give "exact results" can be deceived. For instance, if you have a calculation that is within a rounding error of pi, beyond its resolution, it might be biased to recognize the answer as pi, even though the original expression doesn't strictly equal pi.
Matt Parker has a video on this, where his calculator gives false answers due to near-misses caused by floating point errors. Unless your calculator is a computer algebra system that processes exact expressions properly.
@@carultch Thanks for your answer! Forgot to mention in my original post, but my calculator is not a physical one, but a downloaded scientific calculator app on my phone named "HIPER Calc". Default precision in the pro version is from 15 digits significand and 3 digits exponent as the minimum (compact portrait layout), to 45 digits significand and 6 digits exponent in expanded layout. Decimal precision in fix notation is 6 digits. This can be changed by the user.
In the case of this video, the numeric value displayed for each result expression variant was identical.
skap?
@planetx1595 Thanks for pointing out that typo. Extraneous Norwegian word inserted by autocorrect and now removed.
9^(3/4)
(3²)^(3/4)
3^(3/2)
3^(7/2)
27^(3/2)
(3³)^(3/2)
3^(9/2)
3^(3/2)×3^(7/2)
3^(10/2)
[3^(10/2)]/[3^(9/2)]
3^½ ❤
First time seeing this new exponent law written at the right corner side of the board,we indians didn't know this.....
which law??
√3 ?
If someone doesn't already know how add and multiply fractions, they will certainly not be able to do exponents.
maybe OP is baffled by fractional exponents, like 3^(3/2)
What if the base numerals were all not multiples of three? Would there ever be such a case?
Well, since it's a product you can use a logarithm and the change base formula for logarithms.
logn(x) = logarithm base n of x n^logn(x)=x
logn(n)=1
logn(1)=0
logn(a*b)=logn(a)+logn(b)
logn(a^b)=b*logn(a)
loga(x)=logb(x)/logb(a)
(You can verify this change base formula via: logb(x)=loga(x)*logb(a) x=b^(logb(x))=b^(logb(a)*loga(x))=(b^logb(a))^loga(x)=a^loga(x)=x )
E.g. calculate/simplify: x=(a^b*c^d)/(f^g)
goes like this: x=5^log5(x)=5^y=(a^b*c^d)/(f^g) => y = log5( (a^b)*(c^d)*( (f^g)^-1) ) = b(log5(a)) + d(log5(c)) -g(log5(f)) =>
x=5^( b(log5(a)) + d(log5(c)) -g(log5(f)) )
..and for a=45, d=240 and f=250 this becomes:
y=b + 2b(log5(3)) + d + 4d(log5(2) + d(log5(3))- 3g -g(log5(2)) = b+d-3g + (2b+d)log5(3) + (4d-g)log5(2) => x=5^y=5^(b+d-3g)*3^(2b+d)*2^(4d-g)
And for 4d=g, b+d=3g and/or 2b+d=0 this becomes more simple.
Difficulty level US 😂😂😂
1 👀
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Hello sir. What country are you based my friend, I love you’re teaching style, do you have business contact details?kindly leave below if possible.
The answer is : sqrt(3).
Dudes avatar looks like he's gonna do bad things to that X 👀
There were a couple unnecessary steps in your work, like reducing 10/2 to 5 only to return it to 10/2, but you showed everything and explained all the steps as you went. For someone who is already comfortable with this level of math all the extra steps can be ignored, but for someone who is not already comfortable with this level of math it is good for them to see everything and get an understanding of how and why things work the way they do. By the time you get to this level of math you should have an understanding of fractions, but fractions weren't the point here; the point is about exponents and it's good to see why things work the way they do. Good job to you for explaining everything and not doing so in a condescending manner.
Ezz
Did it in my head in 15 seconds
someone give this man a cookie
@@BenDover69831oatmeal raisin
It's not about speed. It's about explaining all the steps and reason behind them.
Congrats?
Why are you here?
1/9 is the answer according to indian education.....
No you are dumb math is same everywhere. Answer is √3 only.
3/2 = 1.5, 1.5 + 1/2 = 2 not 5
where'd you get 1/2? it was 7/2.
First :)
How???
Comment is older than the video
What how
Teach me
Yoo ,he just changed the date in the settings 😂
This one is easy, people should learn basics more thoroughly. The powers are easy in general if you learn all those rules, basically almost same as normal cases but you first try to make everything have same base and in the end you actually apply the power
idk if the op even did math as a kid, the question is super easy and he's having problem graspt the concept
Op is a kid
@@pedrogarcia8706 doesnt look like it from the way the post was constructed
@@dailydoseofshitpost751 You know that children also can write well, right?
@@dailydoseofshitpost751 Maths ain’t for everyone dude, asking ‘super easy’ questions does not make someone a laughing stock.
chat i only do shitposts on a daily basis, take my comment as a grain of salt
Almost 10 minutes for something I could do in my head in under five seconds when I was 14?! Surely you jest!
That's great but this channel is about teaching things to people who either don't remember or don't get it, etc
Youre so smart u didnt read the name of the channel.
Good explanation. Though I would have kept the 5 exponent at the end as a 10/2 for simplicity. still works. though.
I don't understand how people managed to graduate HS without basic knowledge of simple exponent algebra. Sad.