Math Olympiad | Which number is Larger ? | 90% Failed to solve!
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- Опубліковано 8 вер 2024
- #exponentialproblems
#maths #whichnumberislargertrick
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Another quicker way (perhaps) of solving this inequality problem: We can write 4^5311 as (2^2)^5311 which is 2^10622. Now let us locate the nearest differential values between the two powers of 2 and 5, (their base values) which 128 and 125 (the diff is 3) . That is 2^7 and 5^3. Now 2^7 > 5^3. Let us raise power of 1517 (why?) to both sides. ( Because, we get 2^ (7 x 1517) = 2^10619 which is close to 2^10622 as per our question. We got this value of 1517 by dividing 10622 by 7). Therefore, ((2^7)^1517) > ((5^3)^1517). That is 2^10619 > 5^4551 ... (1). We know 8 >5 or 2^3 > 5 ... (2). So that by logic, 2^10619 x 2^3 > 5^4551 x 5. In other words, 2^ 10622 > 5^4552 > 5^ 4311. Therefore, 2^ 10622 > 5^ 4311 or 4^5311 > 5^ 4311.
thank you Sir, ea eamka value correct
Good work! keep it up !!🎉
Thank you 🙌
( 5^4)^1000 = 625^1000
( 4^5)^1000 = 1024^1000
(x ➖ 3x+2). (x ➖ 3x+2) 5^4311
Within reason, exponents always rule.
4^(3.5)=2^7=128>125=5^3.
4^(5311/4311) > 4^(3.5/3) >5
Luckily it's about maths, and you write it down. That English is incomprehensible.
Use log
Ответ - 0.
Sorry but Your writing is much too small and it is difficult to understand your explanations.
Will try to write big from next time...Thank you for watching!
A ggod mathlete would know the log of 2 and 3!!Just take the log of both sides!!.If I know the log of 2,the log of 5 can easily be found.
A good puzzle solver would know that that is *not* a good way to attack this problem, as not everyone knows approximations to those constants.