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Hello, Like Log, Ln, Sin, Cos, G, g, Pi, etc., e is another decisive and facade number, valid upon Crooks' agreement-the entire ignorant world.

2048 -32=2016. This should be solved in about 10 seconds in our head.

10×9×8×7×6!/6!=x! 5×2×3×3×4×2×7=x! 5×6×3×4×2×7=7×6×5×4×3×2=7!=x! X=7

don't you american know Ruffini's algorithm for god sake?!? Why in the fucking hell do you proceed in that way (extremely difficult and unreliable) after you found the first solution for the equation,which is t = 5 instead of using the Fucking Ruffini's rule?!?

Yes !!! Or ... 2^x = 7^(x + 2) note: log₇(2) = 0.356207 (7^0.356207)^x = 7^(x + 2) 7^(0.356207x) = 7^(x + 2) same base (= 7) 0.356207x = x + 2 0.356207x - x = 2 -0.643793x = 2 x = 2/-0.643793 ■ x ≈ -3.106588 🙂

BS nobody needs in real life 🎉🎉🎉🎉

This is an excellent work! there is another simple method that can be use to solve the given question. See solution to similar question here: ruclips.net/video/ccIRAx0MIfQ/видео.html

mildly overpcomplicated, just take ln of both sides and express x. done.

-14/5

For the exam: ¿Do you have to write down your deduction or is it sufficient to give the solution and the proof 729 + 81 + 27 = 837?

Rational roots theorem gives you x = -2 Divide the expression by x+2 to get x^2 - 3x + 6 and solve for the remaining roots with the quadratic formula. This is how most people who could solve this problem would do it. Your solution is interesting, but looking for those kinds of solutions would probably waste a lot of time on the exam. Especially if the more basic strategies for finding roots of polynomials work well enough.

Love your series, what books would you recommend to someone looking to compete in olympiads?

Perhaps it is because I'm an engineer but I started in my head with trial and error - 2^4 cannot be equal to 1^4 - no point in getting bigger, similarly for a=-1 and below. However 0.5^4 would be the same as (-0.5)^4. That took about 20 seconds of thinking. Do mathematicians always have to take the long route? Runge-Kutta says "guess and answer and then guess a better one"

More faster for my opinion: 3^x is max and <= 837. X=6 3^6=729. 3^y + 3^z= 837-729=108. 3^y max and <=108. Y=4 3^4=81. 3^z = 108-81= 27. 3^3=27. Z=3. Ans. X=6 y=4 z=3

One of the terms must be greater than 1/3 of 837. Only 729 satisfies this, and leaves the sum of the other 2 terms having the last digit as 8. Powers of 3 can only end in digits 3, 7,1 and 9. Therefore perforce the reamaining terms must be 81 and 27.

Got the solution

Obviously a != 0, so 1 = (1-1/a)^4, 1-1/a={1, -1, i, -i}, obviously 1 is dropped which is okay because this is a cubic equation. So 1/a={2,1+i,1-i}, a = {1/2, (1+i)/2, (1-i)/2}.

In Cambridge, UK,it's pronounced zed!

can use polynomial theorem? Horner scheme? btw it's same with log2(5) right?

This is a 30 second mental problem for 8th graders. You need to say which math olympiad it is from. If you cannot do so you should remove the olympiad tag

The problem has infinitely many solutions : choose e.g. x= 0 , y=1 .Then z = ln833/ ln 3 . This way you can construct arbitrarly many solutions.

Write down the powers of 3: 1, 3, 9, 27, 81, 243, 729. From that you immediately see that all of the variables must be less than 7 and exactly one of x, y, z must be 6. Thus you reduced the problem to 3^u+3^v=837-729=98. Similarly you can notice that none of u, v can be can be greater than 4 and exactly one of them must be 4. So subtracting again 3^4=81 we reduce the problem to 3^s=98-81=9. Summing it up all the solutions are permutations of 3, 4 and 6.

Can this be solved without using the W function?

837 = 9*3*31 =(3^3) (27 + 3 +1) = 3^6 +3^4 +3^0 • digits of 837 add to a multiple of 9 so divide by 9 • 3 divides 93, 31 times • break down 31 in powers of 3 • done

Interesting would be if you would allow complex numbers as solutions. What would be the answer then?

Если логарифмировать по основанию 2, то 1/х= log(2)3 x= log(3)2 Два действия за 15 секунд

This takes about 10 seconds by just thinking about the powers of 3.

Это задача 8-9 класса😢

The answer is x = 0,6309

There is more better and simply solution

You don't need to pass you just need to plagerise like the dean.😅

Genial. Gracias-----

Bonne solution .

Radix conversion, usually from base 10 to 2, and 2 to 10, used for all computer calculations, is one of the most common calculations on earth. In this case a radix conversion from 10 to 3 is required. Conversion of n to radix 3. The process is 1. Find the largest 3^(position - 1) < n. 2. Subtract 3. Repeat until remainder 0. Giving 3^6 (729) + 3^4 (81) + 3^3 (27) = n (837).

Interesting video, but if you're going to write the product as 26*5, why not just write it as 25*5 + 5 since you have y^3 + y, so 125+5 = y^3 + y, 5^3 + 5 = y^3 + y therefore y = 5 x = log_2(5).

That one was beautiful👏👏

10!/6!=x! 10•9•8•7=x! 2•5•3^2•2^3•7=x! 2•3•4•5•6•7=x! 7!=x! x=7

Did it a slightly differently, didnt factor out 3^z. Divided through by 3^3, so indices were x-3, y-3,z-3. Observed that indices must be positive, or wouldnt sum to an integer. ie x,y,z>=3. However, if they were all stricly greater than three, you could take a factor of 3 from the whole expression and divide through, which you cant because 31 is prime. Therefore one index must be precisely 0. say x-3=0, x=3. sub in, and subtract 1 from each side and you have the remaining terms sum to 30. But 30 has a factor of 3, so you can divide through to have the indices as y-4,z-4 and equalling 10. Trivially 9 and 1, which yield 5 and 13 respectively.

I was too distracted by the way this person makes an X. Bruh, why you takin the long way? THATS NOT HOW YOU MAKE AN X

-2?

How is it concluded that 3^z = 3^3?

the trick is 130=26*5

Excellent exercise for understanding some the equations with multiple variables. I would not think on the multiple combinations of solutions, this is what I learned today. Thank you

yes 😀👍

You cannot do the step at 2:00 without noting that you are excluding x=0 as a solution.

Such sigma math, so skibidi sus ohio vibes like that 10/10 fanum tax gyat

8^x + 2^x = 130 2^3x + 2^x = 130, let U = 2^x u^3 + u = 130 I’m just gonna use my intuition at this point (if that wasn’t possible I’d just use newtons method). u = 5 so 2^x = 5 so x = log2(5) = ln(5)/ln(2)

My brain: There’s a 5 in here, right? Me: Maybe. But you’re 40 years old, you might have epilepsy and you had an actual seizure 4 weeks ago. My brain: So that’s why I understand absolutely nothing else 😅

Note: this is not a quartic equation. It is a cubic, which is why it has three roots, by the FTOA. It's fastest just to expand the binomial, cancel the a^4 term, find the real root a = 1/2 (you can pretty much do this by inspection or guess and check) and then do simple polynomial division by the factor (2a - 1) to get the quadratic 2a^2 - 2a + 1, which gives you the two complex roots of (1 +/- i)/2..

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