x = y = 16.

If it's very simple to solve then maybe you should've solved it correctly

solution is x = 3, not x! = 3.

Can't cancel (x-2)! And (x-2) The right answer: x(x-1)(x-2)(x-3)!=3(x-1)(x-2) After cancel x(x-3)!=3 If check x=3 that is 3(3-3)!=3 This right (x=3) Can't check x<3 that is road to Gama Function.

You should've written x! as x(x-1)(x-2)(x-3)! and then cancel the (x-1) and (x-2) factors with their corresponding ones on the right side of the equation and that would've left only x(x-3)!=3 wich is only possible if x=3 or, to answer the original question: "find x!", the answer would've been x!=6, because 3!=6.

this guy cannot write the letter x

bro thanks

The cancellation should have left x(x - 3)! = 3. Then you would check for when x >= 3 that solves the problem. (x = 3 is the solution)

I thought this channel was a joke but I guess not.

"x (x-1) (x-2) !" has no clear meaning and non-sense

What? "That's not how this works, that's not how any of this works"

1 is too a answer

3^4 = 81 3^2 = 9 81 + 9?

Hahahah xD what a waste of time

were you never told that 0-equalities are extremely dangerous?

That is some horrendous handwriting.

Just solve these problems algebraically.

I was like "wait, you get x=4x so simply by removing x" but that would be division by zero so wouldnt work The answer is 0 regardless

Pl use a better pen or lazer

math olympiads are getting way too easy. I took 2nd place in 88 in my state and I dont think there was a single question this easy

*4 so 4x = x, 3x = 0, x=0...

How u got more than 16000 views in just one month?

Music name?

You need to state that x and y are whole numbers. The lower bound on x is 13, given that restriction. If you decide to let x and y be non-negative real numbers, you'll get an infinite number of solutions with a lower-bound of x = 4*(log(40))^2/9/(log(2)^2) and y=0.

Merci c compris

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The math is really nice. Thank you

Correct calculation, but much too complicated... but it works like this too... with the approach 3^x =6; x=LOG(3) 6 x= 1.6309

Nice

-10 =-8-2= y^3_2^3+y-2=0 y =2

u aded useless steps

u could use ln insted of log and get it dirctely hhhhhhhh

Faster way would be: set 3^x = 2 * 3, then log base of 3 on both sides. x * log3(3) = log3(2) + log3(3) , replacing log3(3) with 1 gives you the final result x = log3(2) + 1

1 Implicitly connected GROUPS of factors and coefficients are treated as single values throughout mathematics. Distance divided by circumference is equal to revolutions. D÷2πr=D÷πd must be calculated as D÷(2πr)=D÷(πd) to avoid failing dimensional analysis. It can be shown that the viral equations 8"÷2(2"+2")=? And 6"÷2(2"+1")=? Are both, in fact, forms of D÷2πr= revolutions. And must be calculated as 8"÷[2(2"+2")]=1. And 6"÷[2(2"+1")]=1 Or arrive at experimentally, verifiably false results of 16in²and 9in² respectively. Even, or especially, when the radius dimensions are a sum of hub and tread thickness. 120"÷[2π(11"+1")]=120"÷[π(22"+2")]=1.59 revolutions. Volume divided by area to determine depth of contents, V÷πr²=depth must be evaluated as V÷(πr²)=depth, to avoid failing dimensional analysis. 1000ft³÷[π(13'-1')]=2.21' deep Volume divided by spherical volumes V÷4/3πr³=number of flasks filled And must be calculated as V÷[4/3πr³] in order to pass dimensional analysis and not arive at 6 dimensional space in the spurious results. 300in³÷4/3π(3in-0.12in)³= 300in³÷[4/3π(3in-0.12in)³]=3 This is also a good example of how the different division notations are used differently. V÷4/3πr³=V÷[4(1/3)(π)(r³)]=quantity D÷2πr is interpreted as D÷[2πr] a quotient with an implicit grouping of factors and coefficients in the denominator. While D/2πr represents a product of the factors D(1/2)(π)(r) that would not be equal to rotations made by a wheel. In fact, each and every instance of Implicitly joined groups of factors and coefficients must be resolved prior to even acknowledging any other explicit operations in the rest of an equation. To omit this grouping relationship means to arrive at spurious results each and every time. The implicit grouping relationship is inherent in the implicit notation and cannot simply be ignored, without expecting incorrect results. So PEMDAS must be interpreted as Parenthetical expressions(including coefficients) First, also using PEMDAS. Then, Exponents Multiplication Division Addition Subtraction The example problem is in no way excluded from these mathmatical behaviors.