This Math Problem Baffled Everyone | One of the Trickiest Mathematics Questions

[@nevillerichardson-ko8le]

Isn’t it simpler to give the answer as 3 ^9^9

[@rickymort135]

Yes but thats simpler and better but it's not on there. You need to recognise that d is the same as this

[@Felix_Duesenburg]

One should probably add a note that there exists a convention about the order of execution when there are stacked symbols, namely that they are to be worked top-down. Someone who isn't aware of this will stumble over the non-commutative property of exponentiation, i.e. (9^9)^9 != 9^(9^9). The operation under the square root in this example is also known as tetration.

[@europeleeds]

Thanks for this rule. I wondered where the brackets were. √ 9^(9^9)

[@thakurfamily9669]

Finally something I UNDERSTAND!!

[@BlueYammer]

Wolfram Alpha 9^(4.5*9^8)-sqrt(9^9^9) =

[@fatroberto3012]

I would refuse to answer the question on the grounds that the questioner should have used parenthesis to clarify whether they meant 9^(9^9) or (9^9)^9. It's not good enough to assume that any student should know that the first interpretation is correct. Math tests should not be memory tests. There are plenty of other subjects for that.

[@afterthesmash]

It's not a memory test. (a^b)^c is never written that way, because it more usefully reduces to a^(bc). Thus when you see a^b^c you can assume a^(b^c), which can't be usefully reduced. You have to be pretty advanced to encounter an algebra or a special case where exponential is cheaper or easier than multiplication. To calculate 3^100 mod 7, we can use Fermat's Little Theorem. 3^6 ≡ 1 (mod 7) 3^100 = (3^6)^16 * 3^4 3^100 mod 7 ≡ 1^16 * 3^4 mod 7 3^100 mod 7 ≡ 81 mod 7 3^100 mod 7 ≡ 4 FLT states that a^(p-1) ≡ 1 (mod p) if p is prime and a is not divisible by p.

[@oahuhawaii2141]

Learn the precedence rules.

[@oahuhawaii2141]

d.) 9^(4.5*9⁸) A better form is 3^(9⁹) .

[@BoraheBiches-ew7yh]

I have a doubt. In this video 9 is the base and 9^9 is power. But what if someone calculates with 9^9 as the base and 9 as the power?

[@satiremuch2643]

What does the power of a fraction even mean? like, 10^2 is 10 times 10. But 10^(1/2) is the root and 2^(0.3333...)? What are those notation describing? How can it be multiplied with itself partially?

[@DanDart]

It's a weird concept but partially you can think about rational powers as taking the integer power of the numerator and taking the denominator'th root. A little weirder when the power is irrational but there are continuations you can do

[@DanDart]

Specifically to your question 2^0.3333... means the principal cube root of 2

[@satiremuch2643]

@@DanDartI think that this might be a very good way of conceptualizing it. Thank you for the answer !

[@Patrik6920]

2^(.333...) = 2^(1/3) -> What number mutiplied 3 times is equal to 2 ex 9^(1/2) -> what nuber mutiplied 2 times makes nine, aka root of ... A^(1/N) -> what number multiplied B times makes A, aka N:th root of ex 81^(1/4) = 3 * 3 * 3 * 3 , 3 multiplied 4 times... Thers a relationship with Logarithmic functions... Log base ten (on most calculors), ask the question how many times must the number 10 be multiplied to make the value, in a more general term log can have any base (even fractions) ..by using log base rules we can write for example Log base 3(81) as Log(81)/Log(3) and the answer is 4 ... we need to multiply 3 four times to get 81 Log base rule: LogA(B) = LogX(B)/LogX(A); X is an arbitrary log base ex: 3^B = 81-> B = Log(81)/Log(3) = 4 (4:th root of 81 is 3, or 81^(1/4) = 3) ps. on most caculators it just say 'Log' wich is Log base 10, Log base 2 is also quite common as it relates to binary (two digits 1 and 0)...

[@billj5645]

I think it would be easier to write 9 to the 9th to the 4.5th. If you are going to divide the exponent by the base of the radical then you divide the upper 9 by 2, the other two 9s are unaffected. You could even write it as 9 to the 9th in parentheses and put the last 9 on the outside of the parentheses.

[@oahuhawaii2141]

But that's not how math works.

[@billj5645]

@@oahuhawaii2141 try it with a calculator

[@EnglishByGhMandan]

Is this simplification correct?

[@Carl_Gustav_Häyhä]

4,43426^38 (15 sec) 😀

[@cosumel]

9^9^9^(0.5)=9^9^3. Power towers are evaluated from the top down.

[@rickymort135]

You're right and wrong: Yes the stack is evaluated 9^(9^9), but what you're evaluating is the square root of the whole thing, i.e.: (9^(9^9))^0.5 = 9^(0.5*(9^9)) = 9^(0.5*9*(9^8)) = 9^(4.5*(9^8)) Also 3^(9^9) would work but that's not on there

[@patriciaceli1536]

Nice!!! Gravias!!!

[@WilliamFairfield-qq2tq]

Out of curiosity as someone who can do basic math but always struggled with this type. Where on earth were you taught this? My school did not dip into any of this, ever and I was in in what we called top sets

[@Nudnik1]

Your 9 looks like a g.. Excellent channel Thank you

[@halukguler4995]

Wrong solution!!!

[@johnpaullogan1365]

obviously it is 9^((9^9)/2). simple enoug. the answers are also obviously in ascending order. 9^3 is obviously less than (9^9)/2. answer is d

[@robertveith6383]

Write sentences.

[@randomstuff8828]

I got *6.54312443084912643965856048090844034127836565176422210... × 10^184846549*

[@ZeroGravityDog]

Why write 'g' if you mean '9'?

[@thomasg5554]

3⁸¹

[@oahuhawaii2141]

Wrong. You have the precedence rule wrong.

[@JulioVerne-p6w]

@Higher Matematics The result is inaccurate, 3^81 is correct. I did the test and it is correct!!!!!

[@ZeroGravityDog]

No it isn't!!!!!. If you square 3 ^ 81 you get 9 ^ 81 which is 9 ^ ( 9 * 9 ) which is a tiny tiny splinter compared to 9 ^ ( 9 ^ 9 ). What was the test by the way?

[@tambourineman17]

I get the same answer. 9^(81/2) = 3^81

[@ZeroGravityDog]

@@tambourineman17 Where does the 81 come from? 9^9 is a lot bigger than 9x9

[@nazneenakhtar2322]

You multiplied 9 with 9 but that is actually 9 to the power 9

[@oahuhawaii2141]

@ZeroGravityDog: The ones who have the same wrong result got the precedence rule wrong; they used (9⁹)⁹ = 9⁸¹, and took the square root.

[@vanshkansal73]

This question is nothing compared to what i have solved, i have solved much much more difficult questions in my regular 12th standard class while preparing for JEE ADVANCED

[@afterthesmash]

It doesn't matter what you prepared for, if you square each possible answer you quickly find the right one. The video is kind of stupid for not spotting that.

[@AkshayMohabeyAR]

Lol, this is just lame 😒