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AminusB
United States
Приєднався 18 гру 2023
Welcome to AminusB, the place to be for all things math! We create short, engaging videos that explore the beauty and wonder of numbers and mathematics
Our channel covers a wide range of mathematical topics, from puzzles and brain teasers to advanced concepts like calculus and statistics. We also provide video lessons on high school math topics such as algebra and geometry, perfect for students looking to supplement their learning or anyone who wants to brush up on the basics.
Our goal is to make math accessible and enjoyable for everyone, so we're always looking for new and exciting ways to present mathematical ideas.
So if you're ready to dive into the world of math and explore its beauty and complexity, make sure to subscribe to our channel and join the AminusB community today!
Subscribe: youtube.com/@AminusB7?si=bSr_8ra8gqpnkh_9
Our channel covers a wide range of mathematical topics, from puzzles and brain teasers to advanced concepts like calculus and statistics. We also provide video lessons on high school math topics such as algebra and geometry, perfect for students looking to supplement their learning or anyone who wants to brush up on the basics.
Our goal is to make math accessible and enjoyable for everyone, so we're always looking for new and exciting ways to present mathematical ideas.
So if you're ready to dive into the world of math and explore its beauty and complexity, make sure to subscribe to our channel and join the AminusB community today!
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A Logarithm Power Tower | How to solve?
A nice math Olympiad question to solve
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The 9 looks like “の” for Japanese people
Thanks sir ❤
You wrote the wrong equation.
Ln(e^2)*Ln(e^2)*Ln(e^2)=8 Input log(e^2) log(e^2) log(e^2) = 8 Result True Reduced logarithmic form True Exponential form e^(2 (2 2)) = e^8
Ln[x]^Ln[x]^Ln[x]=8 x≈6.56763616490964541630441144595252656019954099174055231464219082987772003108251185989637969840025566076916778677217551999998909646609020045329611392356463775295113438392880449734625839929883801086530590966026988410714568258323183871056024416925615962889006706840999251686477918385904129494470744372670650532194693588245703507585555039403235493609568757345514370533248858499047718122515073141345468830976835004933569574306186179922180284280884689222694463569577580127540040753579111794493820542413468742953608770787489110160636295653496050684740988350746967526808877685321112904814849128346460422718325931183140611717885974000817241077125025676986182179258662355960207566097117341567331072888989614860434342935934726745918084523876854116599178093510371253665923464609017153...
...Do not forget, please, when you take the roots, how should put a plus and a minus on the right side...
...In this minus one it is the second answer...
OMG IS THIS REAL I THOUGHT IT WAS 3 😭😭😭
Is -1 answer too?
Yes. Thanks.
Si lo hubiera resuelto como una diferencia de cuadrados, hubiera encontrado la otra solución que es 17
(x-1)³=64×27 x³-3x²+3x-1=1728 x³-3x²+3x-1729=0 (x-13)(x²+10x+133)=0 (x-13)(x+5+6i(sqrt3))(x+5-6i(sqrt3))=0 😊
I didn’t use a calculator I did it in my head.
(2^33+2^22+2^11)/(2^33-1)=1.00048851978505129457743038593063019052271617 recurring=2048/2047=1 1/2047
sq root=3+2i where i^2=-1(complex)
Nice ❤
Good
Simple but excellent and nice thanks
Thanks for your comment
I am not a mathematician, I am actually trying hard to learn at this current moment and I'm under pressure. However, I took root6 of 3^6 = (x/2)^6 which gave me 3 = x/2 then 3 x 2 = 6 therefore x = 6 Easy.... did this in my head essentially (are you using a more complex method specifically for maths Olympiad?)
this solution includes two special tricks. The first is in recognising the factorisation to the form x= ab/(a+b) the second is in manipulating. that equation into the form 1/x =1/a+1/b An alternative to that approach is to multiply the top and bottom by the conjugates a'=sqrt5-sqrt3 b'=sqrt7-sqrt5 note that aa'=5-3=2 simikarly bb' x=aa'bb'/(aa'b' + bb'a') x=4/[2(b'+a')] x=2/(b'+a') x=2/(sqrt7-sqrt3) rationising the denominator x=2(sqrt7+sqrt3)/(7-3) x=(sqrt7+sqrt3)/2 Answer
Actually, there are 3 answers 🤓☝️
3
sympa...
3/2
3/2
Sqrt[2]/2+Sqrt[2]/2 i=1/Sqrt[2]+1/Sqrt[2] i
Surd[-1,4]=Sqrt[2]/2+(Sqrt[2]/2)i
Surd[-1,4]=0.5Sqrt[2]+0.5Sqrt[2]i
3
Number A ?? Letter you mean
B)
You squared the equation which introduces another extraneous solution. The only answer to this question is +√2 Example: if x = 3 (and only 3) then x^2 = 9 but you can’t just take the square root and say now x = ±3 You squared the equation which introduces another extraneous solution.
Nice solution
b
D
3
b
B
Easy
Простую задачу настолько усложнять😂
Too too hard. =). There is much simpler way exists
❤❤
❤❤
0×5=0