@@sirisaacalbertmravinszky2671 it doesn't make it 2 numbers, but it *does* make it 2 answer choices. Many multiple choice tests have 1 correct answer choice per question, and on such a test you could eliminate c and d for being the same
That's not true at all though..c and dcould still be smaller than b and d..im curious why you didn't realize this? Just wondering becaise you have to show 2 ^36 is smaller than 3^21..
a) c) and d) are all powers of 2 so those are easy to compare in my head. 3^21 = 2^(21*L) where L = log base 2 of 3 which is about 1.585 so that one is the smallest since 3^21 is about equal to 2^33.3. The other 3 choices are higher powers of 2. Many people know 1.5 powers of 2 is about 2.828 so even if you approximated 3 as being 1.6 powers of 2, you would still get the right answer, as 21*1.6 = 33.6 which is still fewer powers of 2 then the runner up which is 36 powers of 2. This can all be done in your head rather easily.
It was easy to see that a > c=d. So a cannot be smallest. Since c=d, neither c nor d can be smallest. Therefore, assuming that the problem has an answer, the only possibility left is b.
Comparing 3^21 to 2^36, we have see there's a common factor of 3 in the exponent, so take the cube root. Thus, we compare 3^7 to 2^12, which we know as 2187 and 4096, respectively. Therefore, we know 3^21 is the lower value, which is choice b).
Since only one of the answers are correct, and two of them are equal, then the answer can only be the last one - which I didn't realize until after you pointed out that B and D are equal (I knew it but it didn't register).
typical mathematician! using proofs to provide a concrete answer to a multiple choice! joking aside, that was a neat trick to change bases to show slower growth with high exponential terms.
3^21 = 3^(3x7) and 2^36=2^(3x12) so take the cube root from both sides and the question becomes: which is smaller 3^7 or 2^12 i.e. 3*729 or 4*1024. 3 3^7 is smaller --> 3^21 is smaller. so answer b is the smallest of these numbers.
@@Pi-zo3gg looking at them you’ll notice c equals d and both smaller than a, since we’re looking for the smallest ONE, so, it can’t be a since it’s already bigger than c and d, it can’t be c or d either because then we’ll have two instead of one. so, b is the only possible one left.
a) 2^40 b) 2^36 c) 3^21 d) 2^36 So, it's between b and c. (and d). Now, 3 = 2^log2(3) then c) = 2^log2(3)*21, about 2^33.3 The smaller number is c Checking a) is about 1.1*10^12 b) is about 6.9*10^10 c) is about 1.0*10^10 d) is equal to b. Edit:got confused between b and c. The smallest number is B
To be a spoilsport, once you know c and d are equal and smaller than a, it must be b.
No problem. I felt it necessary to prove it was b in the video though.
Well, c=d could be the smallest number, why not? Just because it's written in two different ways doesn't make it two numbers.
@@sirisaacalbertmravinszky2671 it doesn't make it 2 numbers, but it *does* make it 2 answer choices. Many multiple choice tests have 1 correct answer choice per question, and on such a test you could eliminate c and d for being the same
That's not true at all though..c and dcould still be smaller than b and d..im curious why you didn't realize this? Just wondering becaise you have to show 2 ^36 is smaller than 3^21..
@@leif1075 Let's assume that there is only 1 correct answer choice, and let's also assume that you know c and d are equal and smaller than a. (
Same method, but pick 27 and 32, so you have 3^21 = (3^3)^7 = 27^7 and 2^36 = (2^5)^7.2 = 32^7.2.
Obviously 3^21 must be smaller.
a) c) and d) are all powers of 2 so those are easy to compare in my head. 3^21 = 2^(21*L) where L = log base 2 of 3 which is about 1.585 so that one is the smallest since 3^21 is about equal to 2^33.3. The other 3 choices are higher powers of 2. Many people know 1.5 powers of 2 is about 2.828 so even if you approximated 3 as being 1.6 powers of 2, you would still get the right answer, as 21*1.6 = 33.6 which is still fewer powers of 2 then the runner up which is 36 powers of 2. This can all be done in your head rather easily.
That’s a great problem! Like the comparison of 4 values and didn’t really think to do it with a decimal exponent
It was easy to see that a > c=d. So a cannot be smallest. Since c=d, neither c nor d can be smallest. Therefore, assuming that the problem has an answer, the only possibility left is b.
Perfect reasoning!! I did likewise
Comparing 3^21 to 2^36, we have see there's a common factor of 3 in the exponent, so take the cube root.
Thus, we compare 3^7 to 2^12, which we know as 2187 and 4096, respectively.
Therefore, we know 3^21 is the lower value, which is choice b).
Thank you very much. l liked the clarity of your reasoning and explanation very much.
Since only one of the answers are correct, and two of them are equal, then the answer can only be the last one - which I didn't realize until after you pointed out that B and D are equal (I knew it but it didn't register).
You mean C and D.
Great video, i really enjoyed
Thanks.
typical mathematician! using proofs to provide a concrete answer to a multiple choice! joking aside, that was a neat trick to change bases to show slower growth with high exponential terms.
3^21 = 3^(3x7) and 2^36=2^(3x12) so take the cube root from both sides and the question becomes: which is smaller 3^7 or 2^12 i.e. 3*729 or 4*1024.
3 3^7 is smaller --> 3^21 is smaller. so answer b is the smallest of these numbers.
yes.
Since c and d are equal to each other (simple to see) and smaller than a), it has to be b) Done
Yes. That’s also my way.
Agree, logic over brute math
Take these thumbs up, my fellow logicians.
Not quite sure why it logically has to be b on inspection. Can you explain please. Thanks.
@@Pi-zo3gg looking at them you’ll notice c equals d and both smaller than a, since we’re looking for the smallest ONE, so, it can’t be a since it’s already bigger than c and d, it can’t be c or d either because then we’ll have two instead of one. so, b is the only possible one left.
a) 2^40
b) 2^36
c) 3^21
d) 2^36
So, it's between b and c. (and d).
Now, 3 = 2^log2(3)
then c) = 2^log2(3)*21, about 2^33.3
The smaller number is c
Checking
a) is about 1.1*10^12
b) is about 6.9*10^10
c) is about 1.0*10^10
d) is equal to b.
Edit:got confused between b and c. The smallest number is B
Did process of elimination
2^2=4
2x18=36
4^18=2^36
8=2^3
3x12=36
4^18=8^12 and they are smaller than 2^40
only remaining option is 3^21
nice n easy
Thanks
the final trick is ok, but (i) one can first simplify the problem by taking the 3rd root, and (ii) bases 27 and 32 would have sufficed
Well spotted 👍
A is the biggest
a=2^40=4^20
b=3^21
Hence a>b
It certainly looks like it however many like the ultimate proof.
Bro this question is appearing in UPSC
Same question in upsc 2023 csat
2022
好
asnwer=(b) 3
Don't be dumb, just log it, log2=0,3 and log3=0,47
Same method here; I happen to know log2 and log3 from schooldays, so didn't need to look them up.
You just basically wandered around lost in the woods for a while...not impressive.
I like to explain my proof because not everyone is at the same level. Can you please send a copy of yours so that I can improve my approach. Thanks.