Germany | Can You solve this ? | Math Olympiad X=? & Y=?

Hello My Dear Family😍😍😍
I hope you all are well 🤗🤗🤗
If you like this video about
How to solve this math problem
please Like & Subscribe my channel as it helps me alot ，🙏🙏🙏🙏

[@user-qr7dw4hk6x]

Бог мой, эту мутотень растянули на столько времени и текста. Здесь делов на полминуты. Нет, надо объяснить, что 2×3 равно именно 6 . Тьфу

[@oahuhawaii2141]

I wanted the solution given as: x = 1 + logₙ(3)/logₙ(3/2)*(logₙ(10)/logₙ(6) - 1) ≈ 1.7724741067515... y = 2 - logₙ(2)/logₙ(3/2)*(logₙ(10)/logₙ(6) - 1) ≈ 1.5126231021869... Where n is from { e, 1.5, 2, 3, 5, 6, 10, 15, 30, ... } . He can do manual calculations with n = 10 , and use log₁₀(2) ≈ .30103 and log₁₀(3) ≈ .47712 .

[@cachotrelles4715]

Excelente explicación, aún sin tener idioma común, ESTÁ EL IDIOMA Y COMPRENSIÓN DE LAS MATEMÁTICAS Y SUS REGLAS 👏👏👏👏👏

[@TheSimCaptain]

It was simpler to just substitute the value of x in the second equation. x + y = 3.285 y = 3.285 - x y = 3.285 - 1.7726 y = 1.5124

[@oahuhawaii2141]

Rearrange the factors in the equations and take the log to produce a system of linear equations, which can be solved quickly via matrix math: 2ˣ * 3ʸ = 18 = 2*3² 3ˣ * 2ʸ = 20 = 2²*5 { x term left, y term right } Take log. Since this presenter likes to expand and calculate with log estimates of small integers, use a = log(2) , b = log(3) , c = log(5) : x*a + y*b = a + 2*b x*b + y*a = 2*a + c [Note: Unicode lacks subscripts for 'c' and 'y'.] Write the 3 matrices of the coefficients, variables, and constants in Mc * Mv = Mk as: ┌ ┐┌ ┐ ┌ ┐ │a b││x│ │a + 2*b│ │b a││y│ │2*a + c│ └ ┘└ ┘ └ ┘ Recall Cramer's rule to evaluate 3 determinants to find x = Dx/D and y = Dy/D . Write x and rearrange for factors of linear terms: x = (a² + 2*a*b - 2*a*b - b*c)/(a² - b²) = (a² - b*c)/(a² - b²) = (a² - b² + b² - b*c)/(a² - b²) = 1 + b*(b - c)/(a - b)/(a + b) x = 1 + b*(c - b)/(b - a)/(a + b) { use factors > 0 } Write y and rearrange for factors of linear terms: y = (2*a² + a*c - a*b - 2*b²)/(a² - b²) = 2 + a*(c - b)/(a - b)/(a + b) y = 2 - a*(c - b)/(b - a)/(a + b) { use factors > 0 } Rewrite with k = (c - b)/(b - a)/(a + b) : x = 1 + b*k y = 2 - a*k A point to note is that c can be rewritten as: c = log(5) = log(10/2) = log(10) - log(2) = c' - a , where c' = log(10) . [The base of log is generic in our solution; it can be e, 2, 3, 5, 6, 10, 15, 30, ..., but 10 is very useful.] Thus, k can be rewritten as: k = (c' - a - b)/(b - a)/(a + b) = (c' - (a + b))/(b - a)/(a + b) = (c'/(a + b) - 1)/(b - a) Substitute for a, b, c' with log as log₁₀ : k = (log(10)/(log(2) + log(3)) - 1)/(log(3) - log(2)) = (log(10)/log(6) - 1)/log(3/2) ≈ 1.6190310096443... x = 1 + log(3)*k ≈ 1.7724741067515... y = 2 - log(2)*k ≈ 1.5126231021869... FYI, the final results are the same using logₑ or ln : k = (ln(10)/ln(6) - 1)/ln(3/2) ≈ .70313623351880... x = 1 + ln(3)*k ≈ 1.7724741067515... y = 2 - ln(2)*k ≈ 1.5126231021869... If you use a simple 4-function calculator, then choose base 10, since c' = log₁₀(10) = 1 . Thus, you need log₁₀(2) and log₁₀(3) only; if you remember them, skip the log₁₀ lookup table: a = log₁₀(2) ≈ .30103 b = log₁₀(3) ≈ .47712 k ≈ (1/(.30103 + .47712) - 1)/(.47712 - .30103) ≈ 1.6190 543126748... { 1.6190 310096443... } x ≈ 1 + .47712*k ≈ 1.7724 831936634... { 1.7724 741067515... } y ≈ 2 - .30103*k ≈ 1.5126 160802554... { 1.5126 231021869... } I put the original full results and inserted spaces to ease comparison. They're good to 5 sig figs.

[@MsNancy1212]

How did you get 1.285

[@elmer6123]

Use natural logs for a straight-forward answer x = (ln3*ln20 - ln2*ln18)/(ln3*ln3 - ln2*ln2) = 1.77247... y = (ln20 - ln3*x)/ln2 = 1.51262...

[@oahuhawaii2141]

For convenience, you can use any base, including 2, 3, 5, 6, 10, 15, 30, and e . If you're doing manual calculations, then base 10 is good, since you can factor 18 and 20 into products of integer powers of 2, 3, and 5. Their log values are the sum of simple log values. Many folks can skip the log lookup table with estimates: log₁₀(2) ≈ .30103 log₁₀(3) ≈ .47712 log₁₀(5) = 1 - log(2) ≈ .69897

[@PedroOrtiz-sh8hs]

excellent exercise

[@АндрейПергаев-з4н]

Есть замечательная формула и решение будет быстрее на 10 минут А именно если с=a^b, то b= log(a) c

[@oahuhawaii2141]

c = aᵇ logₓ(c) = b*logₓ(a) , x > 0 , x ≠ 1 { e, 10, 2, ... } b = logₓ(c)/logₓ(a) For x = a: b = logₐ(c)/logₐ(a) = logₐ(c)/1 = logₐ(c)

[@christianrenemoreau4612]

Remarquons que les valeurs proposées sont des valeurs approchées et le symbole d'égalité EST donc INCORRECT d'un point de vue rigoureux.

[@oahuhawaii2141]

≈

[@naturalsustainable6116]

I think the answer is not correct. When we substitute the x,y value to the original equations, 1st equation : I get 8.6847 , not 18 2nd equation : I get 9.8635 , not 20

[@oahuhawaii2141]

You added the 2 powers instead of multiplied. The equations are: 2ˣ * 3ʸ = 18 3ˣ * 2ʸ = 20 But you calculated: 2ˣ + 3ʸ = ? 3ˣ + 2ʸ = ?

[@wbell539]

Equations of this form: a^xb^y = c can be resolved into linear equations: y = log(a^(-x)c)/log(b). Then you can solve pairs of them.

[@oahuhawaii2141]

The pair of linear equations is: x*log(2) + y*log(3) = log(18) x*log(3) + y*log(2) = log(20)

[@АйратСаляхутдинов-я4щ]

😂3*285

[@luisalfredonarvaeznarvaez5125]

Bello desarrollo de la ecuación

[@porthorcharlee5151]

Incorrect