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I know the powers of two instantly from direct memory up through 2^16, but I usually then have to calculate a bit. I've just never bothered to learn the higher powers. I spent my career working in the digital hardware area, but by the time powers higher than 65536 started to become "more relevant" I had moved on to other facets of the work, so those beyond 16 just never got "burned in" like the lower ones did. I can give them all to you in hexadecimal, though. 🙂 0x40000. There's a reason we made so much use of hex.
for computer person - after you get to 2^18 = 2^10 ( 1K ) * 256 - multiply that - easier than all that stuff after 2^18 in this video. Or 2^18 =- 2^20 (1M ) /4. both give the correct answer and easier.
Before watching: Alright, if you have like a^b^c^d^e^f^g, you'll typically calculate f^g first, then e^(f^g), then d^(e^(f^g)), and so on,. With that in mind, we can ignore the outermost 3 exponents we see. Why? Because those last 3 are 1^(5^9)), and 5^9 is a real number, and 1^X = 1 for real X. So what this actually meansi s that we're looking at √ 2^(6^2). But square root is simply an exponent of 1/2. Thus, √(2^(6^2)) = (2^((1/2)6^2)) = 2^(36/2) = 2^18. Solution is 2^18. Because we're not solving for X, but simply evaluating a square root, we do not include the negative root in this case. Most math classes will allow you to leave the term 2^18 as-is. However, if your class is currently doing a unit on *evaluating* exponents, you may have to multiply it out. Quickest way without knowing the higher powers of 2 (suppose you only know as far as...2^12 = 4096) would be to do (2^9)(2^9) = 512*512 = 1024 + 5120 +256000 = 262144 Honestly 4096*64 is probably going to go faster, but I went with what popped into my head.
I think this problem is more interesting if instead of solving for the square root of 2^6^2^1^5^9 you solve for the square root of 2^6^2^1^4^4 which has an answer with same digits as the problem
Forget powers of 1: result is always 1. 2 power 1 is 2. 6 power 2 = 36. But you are in a square root, and a square root of any number power 2 is this number. Then, square number of 2 power 36 is 2 power 18. Then, the result is 2 power 18 = 512 x 512 Piece cake!
@@johnnyragadoo2414 ln(1+2+3)=ln(1)+ln(2)+ln(3) is a singularity, not a démonstration ln is associative. 2^4=4^2 is also a singularity, not a proof power is commutative
@@johnnyragadoo2414 back to the question, you have a 1 in the ladder of powers, therefore you must ignore what comes after. 2^6^2^1= 2^6^2=2^36. Apply square root, 2^(36/2)=2^18=262144
@@johnnyragadoo2414 You must read power from right to left 1 ^ 5 ^ 9 = 1 ^ (5 ^ 9) = 1. Result 262144. Period. And if you want to change the rules of math, try with prime numbers and figure out (2^3)^5 and 2^(3^5) are not the same. Result 262144
Did you do it right? You work from the top down. You'd calculate x = 5^9, and then 1^x. But that will just be 1, because 1^ is 1. Then we have 2^1, which is 2. Then 6^2, which is 36. Then 2^36. Then we take the square root of the whole thing, which is 2^18. You do not start at the bottom 2 and work your way up. The answer is 2^18.
Wrong-you perform successive exponentiations from the top down. 2^6^2 should be interpreted as 2^(6^2), or 2^36. The square root of that is then 2^(36/2) = 2^18.
@@ThreePointOneFou Are you saying that 2^6^2 is not the same as 2^12? Even in this video a^b^c is represented as the same as a^(b*c). Evaluate exponentiation from the bottom up. Otherwise 2^6^2 is 2^36 and not equal to 2^(6*2).
@@johnnyragadoo2414 Successive exponentiations are supposed to be evaluated from _right_ to _left_ when written horizontally, to reflect the top-down rule that applies to exponentiation. (Ideally, exponentiation wouldn't be written like that, but those are the limits we're stuck with in commentbforums without mathML or similar support.) The problem shown should be evaluated as though it were written 2^(6^(2^(1^(5^9)))).
@@ThreePointOneFou Then 2^6^2^1^5^9 wouldn't equal 2^(6*2*1*5*9). As it shows in the video, a^b^c = a^(b*c). So, which is it? Should a^b^c = a^(b*c)? I think it should.
Correct, 2^6^2^1^5^9 is not equal to 2^(6*2*1*5*9). And no, a^b^c should be evaluated as a^(b^c). The video does everything correctly, apart from a bit of slightly sloppy writing when showing the rule that (a^b)^c = a^(b*c). When evaluating exponents, the exponent must be fully determined before raising the base to that power-in other words, 6^2^1^5^9 must be calculated _before_ raising 2 to that power. Thus, 2^1^5^9 must be calculated before raising 6 to _that_ power, meaning 1^5^9 must be evaluated before raising 2 to _that_ power, etc. In other words, exponents must be calculated from the top down, or from right to left in horizontal notation.
No !!!!!!! Quan fas (1⁵)⁹, dius que és igual a 1 i no és cert !!!!! Tractes l'1 com si fos Base i no ho és -----> és un exponent!!! No és correcta El correcta és V(((((6)²)¹)⁵)⁹)= V6⁹⁰= 6⁴⁵
