Germany l can you solve this?? l Olympiad Math exponential problem

Рет қаралды 105,505

Math Master TV

Күн бұрын

Пікірлер: 55

@nikhileshkabiraj5613 Ай бұрын

Supper, easily understand a lot of thanks sir.

@pamnetwork7787 Ай бұрын

Nice explanation,good handwriting

@sergedd Ай бұрын

take x=2^(a) then x+(1/x) = 4 then x^2 + 1 - 4x = 0 resolve x = 2 + sqr(3) or x = 2 - sqr(3) But = x=2^(a) -> a=(ln(2+sqr(3)))/ln(2) Or a=(ln(2-sqr(3)))/ln(2)

@dmk_5736 23 күн бұрын

a=±log₂(2+√3) p.s.log₂(2+√3)+log₂(2-√3) = log₂(4-3) = 0, which also obvious from symmetry of initial equation around zero.

@bernardbaz2003 7 күн бұрын

A more interesting one would be 2^(a) +2^(-a)=1 (no real solution)

@nigamroy 2 ай бұрын

Let 2^a =m Therefore m+1/m=4 m² - 4m +1 =0 Apply Sridhar Acharya equatio m =( 4±√16-4)/2 m= (4±√12)/2 m=(4 ± 2√3)/2 m=2 ± √3 2^a = 2±√3 So a = log(2+√3) of base 2. and a = log (2-√3) of base 2.

@mertuderse144 4 күн бұрын

I thought like that also 😂😂😂 he just make that harder and longer

@قدرتاللاهعظیمیعلمداری Ай бұрын

Thanks , that was interesting

@charlysanabria3935 15 күн бұрын

Very well. Thanks.

@b213videoz 20 күн бұрын

let x = 2^a x + 1/x = 4 x² + 1 = 4x x² - 4x + 1 = 0 D = 4² - (4 * 1 * 1) = 16 - 4 = 12 x = +/- [4 - sqrt(12)] / 2 Do the rest, I can't be bothered with this tedious drab - typing math formulas on the smartphone while taking a bath is painful 😂 but long story short it boils to: 1) x = 0.26795; 2^a = 0.26795; a = log(0.26795) / 2 2) x = 3.7321 ...do the same, I'm bored 😊 6:03 this was clever...Like

@marcgriselhubert3915 16 күн бұрын

2^a + 2^(-a) = exp(a.ln(2)) + exp(-a.ln(2)) = 2.Ch(a.ln(2)). So the given equation is equivalent to Ch(a.ln(2)) = 2. It is possible as 2 >= 1 and has two solutions: a.ln(2) = Argch(2) = ln(2 + sqrt(3)) or a.ln(2) = -Argch(2) = -ln(2 + sqrt(3)), (Remind that Argch(x) = ln(x + sqrt(x^2 - 1)) for x>=1) so finally a = ln(2 + sqrt(3)) / ln(2) or its opposite. (This opposite is also equal to ln(2 - sqrt(3)) / ln(2) because ln(2 + sqrt(3)) + ln(2 - sqrt(3) = ln(1) = 0) (The solution ln(2 + sqrt(3)) / ln(2) can also be written log2(2 + sqrt(3)) )

@Qwentar Ай бұрын

Sometimes with fractions like those, it's easier to multiply the one fraction by 1, in the form of the conjugate of the denominator for a difference of squares. Example: 1/(2+sqrt(3)) isn't nice, we'd rather have a single number in the denominator, not a composite number with square roots. Multiply it by (2-sqrt(3))/(2-sqrt(3)); this is effectively multiplying that one number / fraction by 1 which has no effect on the rest of the equation. You get (1*(2-sqrt(3))/((2+sqrt(3))*(2-sqrt(3))), which gives you (2-sqrt(3))/(4-3)

@قدرتاللاهعظیمیعلمداری Ай бұрын

@@Qwentar thank you , for your explanation.

@electricalethertv5041 5 күн бұрын

Из соображений симметрии, если a - решение, то -a - тоже решение. Таким образом, аффтар не нашел половину решений, следовательно задача не решена. Корректное решение: а=±log₂(2±√3)

@luiscantilloalvarado8444 6 күн бұрын

❗❗❗❗😃😃💪💪

@RafayKhan-h6q 2 ай бұрын

🎉🎉🎉🎉❤😮

@carlosa.correia7938 Ай бұрын

beautifull

@momoz74 12 күн бұрын

Cool problem, but I have a complaint. Last part is unnecessarily complicated. By definition of the logarithm if 2^a= x then a= log (2) of x. If 2^a = 2+sqrt3 then a = log (2) of (2+sqrt3).

@mohanpachalag3944 Ай бұрын

1 / (2+√3) = [( 2- √3)/ ( 2+√3) (2-√3)] multiplying by conjugate = (2 -√3)/(4-3) = (2 - √3 ) (2 +√3 ) + 1/ ( 2 + √3) = (2 +√3) + ( 2 - √3) = 4

@alextsang1205 Ай бұрын

Use calculator to “test” the values

@AndrewLyu-fu8wm 11 күн бұрын

What intrigues me is that if such mathematical equations can be found being reflected in real world?

@destamengesha5616 Ай бұрын

talent showing the method solving the problems.

@rolandmengedoth2191 26 күн бұрын

Well done, but nothing for daliy Business math😊😊

@user-yk4bf8vi8o Ай бұрын

Музыка хорошая, приятная такая. Спасибо

@Teachbajrangi 20 күн бұрын

Write= properly

@1234mxvzdakoyhgftew 2 ай бұрын

Glory to the victory!

@davidshen5916 Ай бұрын

Y=2^A， 2^-A=1/Y， Y+1/Y=4， Y^2-4Y+1=0， Y=2+-Sqrt（3）

@boeubanks7507 7 күн бұрын

Umm, it's a bit sloppy at the end. You just assume that the second solution works. That is not guaranteed. It needs to be checked.

@fetibayat164 Ай бұрын

2üzeri a ya x dersen ikinci derece denklemden kökler 2 eksi 2 kök 3 ve 2 artı 2 kök üç olur 2 üzeri a eşitliğini yap daha kısa sürer

@aristotelesbenicio7476 15 күн бұрын

too long. why not use base 2 on log directly???

@ludmilak9396 2 ай бұрын

У меня картинка не работает. Вот такие ответы? а1,2 = log по основанию 2 аргумента (2+ -- (3)^(1/2))

@ENOSKASHIMBA-h7u 11 күн бұрын

17 minutes for that simple quetion? give us another method

@raf7037 29 күн бұрын

Must write X1/2,because is binom

@jenjr803 Ай бұрын

好詳細的解法，但解一題需要將近20分鐘，我還是放棄好了

@shazhu2455 23 күн бұрын

Why don’t you just take log of base 2 in the first place?

@And-Los 18 күн бұрын

The correct answer correct answer is: 1) log2(2+sqrt(3)) 2) log2(2-sqrt(3)) 3) -log2(2+sqrt(3)) 4) -log2(2-sqrt(3)) LOOOSER!!!!!

@ももも-h6b 11 күн бұрын

you should tell 2^a not equals zero

@ericasi4562 25 күн бұрын

Неужели обычное квадратное уравнение относительно 2^а достойно быть в разделе олимпиады по математике? Это же школьная программа по алгебре 10-11 класс "показательные уравнения".