@BobbyPartson Are you sure? I am sure I can reduce two numbers by a proportionate amount and still preserve the relationship between them. Let's try it with something more obvious... Clearly 2^10 < 3^10 But 2^10 = 2^5 × 2^5 And 3^10 = 3^5 × 3^5 So 2^5 < 3^5 In other words, I can divide both exponents by 2, and still maintain the original relationship, even when the bases are different.
@@BobbyPartson it works. Take the logarithm of both sides, which is continuous and monotonic increasing so it preserves order. Then bring down the exponent and divide both sides by 12. Then exponentiate both sides to undo the logarithm leaving 6^3 and 4^4.
6^36= 6^6*6 4^48= 4^8*6 We both have *6 in both so he can have 6^6 amd 4^8 6^6=(2*3)^6 4^8=4*4*4^6= (2*3)6 and (4*4)6 6^6 and 16^6 And 16 is bigger. It's basically the same,but its equal,cause you dont get big numbers with this
Personally I would plug one in my calculator and subtract the other number from it. If you get a negative, the second number is larger. If you get a positive number, the first one is larger. So you don't have to do any math or even look at the number, just the sign in front of it.
Very easy, we have a test in Saudi Arabia called the "Qudrat" test or the aptitude test. This test is almost impossible to get a perfect score, and if you get it, you will be honored and enter competitions that will get you $10,000 as a minimum.
Eh it’s a 50% chance you get it right
so ur saying that this method doesn't work for all?
@@staybased9320no they are just too lazy to work it out and will just guess which number is bigger
@@staybased9320omg is that a strawman fallacy, that's a rare comment pull right there
"Show your work!" Moment
u probably wouldn't get any marks if u don't show any working out
Divide both exponents by 12 and you get:
6^3 < 4^4
216 < 256
that's bad logic because the bases are different.
@BobbyPartson Are you sure?
I am sure I can reduce two numbers by a proportionate amount and still preserve the relationship between them.
Let's try it with something more obvious...
Clearly 2^10 < 3^10
But 2^10 = 2^5 × 2^5
And 3^10 = 3^5 × 3^5
So 2^5 < 3^5
In other words, I can divide both exponents by 2, and still maintain the original relationship, even when the bases are different.
@@BobbyPartson it works. Take the logarithm of both sides, which is continuous and monotonic increasing so it preserves order. Then bring down the exponent and divide both sides by 12. Then exponentiate both sides to undo the logarithm leaving 6^3 and 4^4.
rewrite 6^3 as (2*3)^3 = 2^3 * 3^3 and 4^4 as (2*2)^4 = 2^8, then shorten by 2^3, now it's 3^3 =27 vs. 2^5 = 32
Well on the sats you get Desmos and you can absolutely just type these into desmos so
I think it was pretty blatant that 4 to the 48 would be bigger
🤓👆
@@mrpayday7055 😐
@@MrRayz 🥰
Helpful!
Or you can just strike off the ^12 at step 2, it makes it easier to see if
6^3 < 4^4
Nice mathSexam
You can take the 12 root of them
just take A=6³⁶ and take log to the base 10 on both sides and B=4⁴⁸ and do the same, its way easier if u wanna prove it with maths and not logic
6^36= 6^6*6 4^48= 4^8*6 We both have *6 in both so he can have 6^6 amd 4^8 6^6=(2*3)^6 4^8=4*4*4^6=
(2*3)6 and (4*4)6 6^6 and 16^6
And 16 is bigger. It's basically the same,but its equal,cause you dont get big numbers with this
36 log 6 and 48 log 4 are not too big for a calculator.
Divide both sides by 12 then exponentiate again to undo the logarithm and you can do it by hand.
@@abebuckingham8198 Yeah, but that is exactly what the video did, just with a different writing.
Why is there a "math sex am" tag on the title?
There are so many easier ways to do this
It doesn't get much easier than this, took me like 5s
if you know anything about exponential growth than this is a no brainer xD
And yet 4^46 is less than half of 6^36. The higher power doesn't always overcome the lower base.
Or you just take the logarithm.
6.8913738e+28 diff
If they dont have the same exponent?
nearly halfway through the video, I got the correct answer.
Why can’t we rewrite 6^36 into 6^6*6?
u can bro but its harder to do 6^6 than to do 6^3
Why would we if we can write it as 6^(3*12)?
6^2^6 vs 4^8^6
4^8 is clearly bigger than 6^2
Wrong. 6^36 is not equal to 6^2^6. Correct answer, wrong method.
@@nickcellino1503 it would be 6^(6^2)
@@nickcellino1503lemme just call someone out and not help or correct them rq
Pov: Christians and Chinese people 😱
Multiply 6 by 6 then you are left with 36^35 and then multiply 4 by 4 nine times then you are left with 36^39, now just compare their powers
Решил по теореме "Степень пизже основания"
I got it correct at 10years old
They really are not too big for a calc
Just use calculator
The point is to not use one.
Súper! ❤
NICE
Степень пизже основания
Bro is cooking
You can also just have 6^36 - 4^48 in your calculator. If the answer is negative, then it's 4^48. I know it's no calculator, but why. That's dumb
Personally I would plug one in my calculator and subtract the other number from it. If you get a negative, the second number is larger. If you get a positive number, the first one is larger. So you don't have to do any math or even look at the number, just the sign in front of it.
No, they are not so big for calculators and can be easily calculated using C# BigNumber
This is how I would do this
(4^4)^12
(6^3)^12
256^12
216^12
4^48 is bigger
I put this in desmos
6^36 = 3.838...^48
4^48 = 6.3496^36
Very easy, we have a test in Saudi Arabia called the "Qudrat" test or the aptitude test. This test is almost impossible to get a perfect score, and if you get it, you will be honored and enter competitions that will get you $10,000 as a minimum.
الله يصلحك
بعدين وشهي ذي العشر الألف
ارجع التليجرام بارك الله فيك
@@River-my2sz
جائزة الأمير فيصل بن بندر
ابحث عنها
وفي جوائز و مسابقات كثيرة على الحاصلين ١٠٠ بالقدرات
6 is the answer beacuse1,2,3,4,5,6,
Ever heard of a scientific calculator
every heard of something called using ur brain lil bro
@@backstabber3537 no I’m saying this because he said it’s not easy to do on a calculator and it is
Are they allowed,I mean they should be...?
You can't use a calculator - the numbers would end up too big
@@nateroblox0008no?