Another solution 🙂 Write it as equation x = product 3^(k / 2^k) from 1 to infinity. Then apply the base 3 logarithm to this equation and obtain log₃(x) = sum k (1 /2)^k from 0 to infinty (re-indexing from 1 to 0 does not change the sum). Then evaluate the infinite sum using r/(1 - r)² with r = 1/2 and get log₃(x) = 2 with the final result x = 9.
I came to the same result in a very different way.. I put the initial expression=x. Then I squared both sides. Then: 3*sqrt(9*sqrt(27..))=x^2. I "take out" a 3^2 to each radical, getting 9*(sqrt(3*sqrt(9*sqrt(..)))=x^2 so 9x=x^2 which gives x=9... Is that correct?
Did it like you but the pivotal part I did in reverse - ie multiplying by 1/2 then subtracting. Amazingly I also left it at 2 rather than 3^2 😂😂😂 just realised when I read the comments 😳
Nice.so the result is 9 because the infinite sum is 2 and 3^2=9
?
Yep.
Another solution 🙂 Write it as equation x = product 3^(k / 2^k) from 1 to infinity. Then apply the base 3 logarithm to this equation and obtain log₃(x) = sum k (1 /2)^k from 0 to infinty (re-indexing from 1 to 0 does not change the sum). Then evaluate the infinite sum using r/(1 - r)² with r = 1/2 and get log₃(x) = 2 with the final result x = 9.
You forgot that the result is 3^2=9
You are often hurry to get the result😊
I know
Don't worry, be happy😊
I got 9
3:08 Hmm, Wolfram|Alpha answers 2 for “sum n/2**n from 1 to infinity” query. Therefore infinite radical converges to 3² = 9.
The solution is 3^2 not 2 ?
I think so 😮
You lost me with the r's...😂
very nicely done. bravo!
Thank you very much!
Yes
👏👏👏 beautifull solution!
Thank you! 😊
Thank you Master for your shareing
You are very welcome
I came to the same result in a very different way.. I put the initial expression=x. Then I squared both sides. Then: 3*sqrt(9*sqrt(27..))=x^2. I "take out" a 3^2 to each radical, getting 9*(sqrt(3*sqrt(9*sqrt(..)))=x^2 so 9x=x^2 which gives x=9... Is that correct?
The WolframAlpha syntax for this:
product (3^(n/2^n))
Nice!
It's a test to see if we're paying attention.
Maybe there'll be a follow-up to this.
😮😍
Isn’t it just a limit problem? Lim from 1-♾️ 3^[x/(2^x)]
Did it like you but the pivotal part I did in reverse - ie multiplying by 1/2 then subtracting.
Amazingly I also left it at 2 rather than 3^2 😂😂😂 just realised when I read the comments 😳
😁
nice manipulation
Thanks