Відео

Brilliant... beautiful...

x^6 - 64 = 0 x^(3 * 2) - 8^2 = 0 (x^3)^2 - 8^2 = 0 (x^3 - 8)(x^3 + 8) = 0 (x^3 - 2^3)(x^3 + 2^3) = 0 (x - 2)(x^2 + x * 2 + 2^2)(x + 2)(x^2 - x * 2 + 2^2) = 0 (x - 2)(1 * x^2 + 2 * x + 2^2)(x + 2)(1 * x^2 - 2 * x + 2^2) = 0 Suppose x - 2 = 0 x - 2 = 0 x - 2 + 2 = 0 + 2 x = 2 Suppose 1 * x^2 + 2 * x + 2^2 = 0 1 * x^2 + 2 * x + 2^2 = 0 x = (-2 +/- sqrt[2^2 - 4 * 1 * 2^2]) / (2 * 1) x = (-2 +/- sqrt[1 * 2^2 - 4 * 2^2]) / (2) x = (-2 +/- sqrt[(1 - 4) * 2^2]) / 2 x = (-2 +/- sqrt[(-1) * 3 * 2^2]) / 2 x = (-2 +/- sqrt[-1] * sqrt[3] * sqrt[2^2]) / 2 x = (-2 +/- i * sqrt[3] * 2) / 2 x = 2 * (-1 +/- i * sqrt[3] * 1) / 2 x = 1 * (-1 +/- i * sqrt[3]) x = -1 +/- i * sqrt(3) x = -1 + i * sqrt(3), or x = -1 - i * sqrt(3) Suppose x + 2 = 0 x + 2 = 0 x + 2 - 2 = 0 - 2 x = -2 Suppose 1 * x^2 - 2 * x + 2^2 = 0 1 * x^2 - 2 * x + 2^2 = 0 x = (-[-2] +/- sqrt[(-2)^2 - 4 * 1 * 2^2]) / (2 * 1) x = (2 +/- sqrt[1 * 2^2 - 4 * 2^2]) / (2) x = (2 +/- sqrt[(1 - 4) * 2^2]) / 2 x = (2 +/- sqrt[(-3) * 2^2]) / 2 x = (2 +/- sqrt[(-1) * 3 * 2^2]) / 2 x = (2 +/- sqrt[-1] * sqrt[3] * sqrt[2^2]) / 2 x = (2 +/- i * sqrt[3] * 2) / 2 x = 2 * (1 +/- i * sqrt[3] * 1) / 2 x = 1 * (1 +/- i * sqrt[3]) x = 1 +/- i * sqrt(3) x = 1 + i * sqrt(3), or x = 1 - i * sqrt(3) x1 = 2 x2 = -1 + i * sqrt(3) x3 = -1 - i * sqrt(3) x4 = -2 x5 = 1 + i * sqrt(3) x6 = 1 - i * sqrt(3)

Ah - I forgot to group the X1 and X2 substitutions when finding Y1 and Y2. So my 2sqrt3s had the wrong signs. Nice problem!

3(3^2)^x = 999, (3^2)x = 333, 2xln3 = ln333, 2x=ln333/ln3, x=2.6434

9^x + 9^x + 9^x = 999 3 * 9^x = 999 (1 / 3) * 3 * 9^x = (1 / 3) * 999 9^x = 333 (3^2)^x = 333 3^(2 * x) = 333 3^(x * 2) = 333 (3^x)^2 = 333 sqrt([3^x]^2) = +/- sqrt(333) 3^x = +/- sqrt(3 * 3 * 37) 3^x = +/- sqrt(3^2 * 37) 3^x = +/- 3 * sqrt(37) 3^x / 3 = +/- 3 / 3 * sqrt(37) 3^x * 3^(-1) = +/- sqrt(37) 3^(x - 1) = +/- sqrt(37) 3^(x - 1) = +sqrt(37), or 3^(x - 1) = -sqrt(37) Suppose 3^(x - 1) = sqrt(37) 3^(x - 1) = sqrt(37) 3^(x - 1) = 37^(1 / 2) ln(3^[x - 1]) = ln(37^[1 / 2]) (x - 1) * ln(3) = (1 / 2) * ln(37) (x - 1) * ln(3) / ln(3) = (1 / 2) * ln(37) / ln(3) (x - 1) * log_3(3) = ln(37) / (2 * ln[3]) (x - 1) * 1 = log_3(37) / 2 x - 1 + 1 = log_3(37) / 2 + 1 x = log_3(37) / 2 + 1 Suppose 3^(x - 1) = -sqrt(37) 3^(x - 1) = -sqrt(37) 3^(x - 1) = -37^(1 / 2) ln(3^[x - 1]) = ln(-1 * 37^[1 / 2]) (x - 1) * ln(3) = ln(i^2 * 37^[1 / 2]) (x - 1) * ln(3) / ln(3) = ln(i^2 * 37^[1 / 2]) / ln(3) (x - 1) * log_3(3) = ln(i^2) / ln(3) + ln(37^[1 / 2]) / ln(3) (x - 1) * 1 = ln(e^[i * tau / 2]) / ln(3) + (1 / 2) * ln(37) / ln(3) x - 1 = (i * tau / 2) * ln(e) / ln(3) + ln(37) / (2 * ln[3]) x - 1 = (i * tau / 2) * 1 / ln(3) + ln(37) / (2 * ln[3]) x - 1 = i * tau / (2 * ln[3]) + log_3(37) / 2 x - 1 + 1 = i * tau / (2 * ln[3]) + log_3(37) / 2 + 1 x = i * tau / (2 * ln[3]) + log_3(37) / 2 + 1 x1 = log_3(37) + 1 x2 = i * tau / (2 * ln[3]) + log_3(37) + 1

I'd love to see how you derive that 1st rule. I've never run across it before. Great problem/solution!!

Um exercício trabalhoso mas instrutivo. Bom para testar as propriedades da radiciação.

Nice problem and solution! I grasp what you did to solve the equation. Im just not following how you choose to raise both sides to ^3 at the beginning. A typical student wouldnt necessarily know to use 3. What did you see in the given equation that prompted you to select the 3? Thank you very much.

But there are two complex roots too: m^(m^3) = 729 (m^[m^3])^3 = 729^3 (m^[m^3])^3 = (9^3)^3 m^(m^3 * 3) = (9^3)^3 m^(3 * m^3) = 9^(3 * 3) (m^3)^(m^3) = 9^9 m^3 = 9 m^3 = (cbrt[9])^3 m^3 = (9^[1 / 3])^3 m^3 = ([3^2]^[1 / 3])^3 m^3 = (3^[2 * (1 / 3)])^3 m^3 = (3^[2 / 3])^3 Let n = 3^(2 / 3) m^3 = n^3 m^3 - n^3 = n^3 - n^3 m^3 - n^3 = 0 (m - n)(m^2 + m * n + n^2) = 0 (m - n)(1 * m^2 + n * m + n^2) = 0 m - n = 0, or 1 * m^2 + n * m + n^2 = 0 Suppose m - n = 0 m - n = 0 m - n + n = 0 + n m = n Remember, n = 3^(2 / 3) m = 3^(2 / 3) Suppose 1 * m^2 + n * m + n^2 = 0 1 * m^2 + n * m + n^2 = 0 m = (-n +/- sqrt[n^2 - 4 * 1 * n^2]) / (2 * 1) m = (-n +/- sqrt[1 * n^2 - 4 * n^2]) / (2) m = (-n +/- sqrt[(1 - 4) * n^2]) / 2 m = (-n +/- sqrt[(-3) * n^2]) / 2 m = (-n +/- sqrt[(-1) * 3 * n^2]) / 2 m = (-n +/- sqrt[-1] * sqrt[3] * sqrt[n^2]) / 2 m = (-n +/- i * sqrt[3] * n) / 2 m = n * (-1 +/- i * sqrt[3] * 1) / 2 m = 3^(2 / 3) * (-1 +/- i * 3^[1 / 2]) / 2^1 m = 3^(2 / 3) * 2^(-1) * (-1 +/- i * 3^[1 / 2]) m = 3^(2 / 3) * 2^(-1) * (-1 + i * 3^[1 / 2]), or m = 3^(2 / 3) * 2^(-1) * (-1 - i * 3^[1 / 2]) m1 = 3^(2 / 3) m2 = 3^(2 / 3) * 2^(-1) * (-1 + i * 3^[1 / 2]) m3 = 3^(2 / 3) * 2^(-1) * (-1 - i * 3^[1 / 2])

Here's how I solved the problem: 1. Start with the given equation: 3^(x+4) - 4^(x+3) = 0 2. Rewrite the equation: 3^(x+4) = 4^(x+3) 3. Take logarithms on both sides: (x+4) * log(3) = (x+3) * log(4) 4. Divide both sides by (x+3): (x+4) / (x+3) = log(4) / log(3) 5. Apply componendo and dividendo: [(x+4) + (x+3)] / [(x+4) - (x+3)] = [log(4) + log(3)] / [log(4) - log(3)] 6. Simplify: (2x + 7) / 1 = [log(4) + log(3)] / [log(4) - log(3)] 7. Putting log(4) = 1.3862943611198906 and log(3) = 1.0986122886681096 on the RHS, we get: 2x + 7 = 8.637683358612833 8. Solve for x: x = (8.637683358612833 - 7) / 2 = 0.8188416793064164 Hope this helps! 😊

First observe that 4^y = 3 implies y=ln(3)/ln(4). Thus, 3^(x+4)=4^(x+3) implies 3^(x+4)=3^[(ln(3)/ln(4))*(x+3)]. Hence, x+4 = (ln(3)/ln(4))*(x+3) and finally, x = (3*ln(3)/ln(4) - 4) / (1 - ln(3)/ln(4)). You could simplify or use a base 3 or base 4 log instead of ln. Whatever... I do not believe this is a final year problem at cambridge. It is highschool level maths.

Thank you, but it's much easier writing 3^(x+4)=4^(x+3) and here applying log in both sides.

3^(x + 4) - 4^(x + 3) = 0 3^(x + 4) - 4^(x + 3) + 4^(x + 3) = 0 + 4^(x + 3) 3^(x + 4) = 4^(x + 3) 4^(x + 3) = 3^(x + 4) 4^x * 4^3 = 3^x * 3^4 4^x * 4^3 / 4^3 = 3^x * 3^4 / 4^3 4^x * 1 = 3^x * 3^4 / 4^3 (1 / 3^x) * 4^x = (1 / 3^x) * 3^x * 3^4 / 4^3 4^x / 3^x = 3^4 / 4^3 (4 / 3)^x = 3^4 / 4^3 log([4 / 3]^x) = log(3^4 / 4^3) x * log(4 / 3) = log(3^4 / 4^3) x * log(4 / 3) / log(4 / 3) = log(3^4 / 4^3) / log(4 / 3) x * 1 = log(3^4 / 4^3) / log(4 / 3) x = (log[3^4] - log[4^3]) / log(4 / 3) x = (4 * log[3] - 3 * log[4]) / log(4 / 3) x = (4 * log[3] - 3 * log[2^2]) / log(4 / 3) x = (2 * 2 * log[3] - 3 * 2 * log[2]) / log(4 / 3) x = 2 * (2 * log[3] - 3 * log[2]) / log(4 / 3) x = 2 * (2 * log[3] / log[4 / 3] - 3 * log[2] / log[4 / 3]) x = 2 * (2 * log_base[4 / 3]_of[3] - 3 * log_base[4 / 3]_of[2])

64

The answer is -1/2

A very big request to the author: I would be happy to see a way to solve this problem not in 2:45, but in at least 5 minutes. And if you can provide a solution that requires 10 minutes, then I will insist that you deserve the title of the best methodologist-mathematician in the world.

Халяву решил радуется

Nice video.

Critically not mentioned is that (1-2^½)^2 would also be valid save that 9^½>8^½ hence the order is required or it would give an incorrect result.

Why go to such lengths? You can simply solve it by: 1) simplifying sqr root of 9 to 3, 2) approximate sqr root of 8 to 2.828, 3) subtract the values: 3 - 2.828 = 0.172 and 4) the sqr root of 0.172 is 0.415.

I did only five line calcu and you made 3 more minute video than my lines

Hi, for this types of solutions please do the lcm for the number part, it will get easy for us to understand. For example here you do lcm for 8064 and you will know the result.

The system of equations has no real solutions!

👍

x^2=ln{x*(-5)} x^2=(-5)*ln{x} x^2={5*(-2)/2}*ln{x} x^2={5/2}*ln{x^(-2)} >>> <<< {2/5}=x^(-2) * ln{x^(-2)} {2/5}=exp[{ln{x^(-2)}] * ln{x^(-2)} || W() W(2/5)=W( ln{x^(-2)} * exp[{ln{x^(-2)}] ) W(2/5)= ln{x^(-2)} || exp() exp(W(2/5))=exp( ln{x^(-2)}) exp(W(2/5))=x^(-2) || ^(-1/2) 1/SQRT{exp(W(2/5)}=x But for more clarity x=exp(-1/2*W(2/5))

shouldnt there be +- root at the end

Fantastic! Mathematics is like a language; you need to understand its rules to grasp it fully. In the future, could you explain the rules better as you go? This way, people who aren't familiar with them won't get confused.

X = 10 (10-3)^4 = 7^4. 😎😎😎

Several times you did and then undid the same step. 2^(x+1)=3^x log(2^(x+1))=log(3^x) (x+1)log(2)=xlog(3) xlog(2)+log(2)=xlog(3) log(2)=xlog(3)-xlog(2) log(2)=x(log(3)-log(2)) x=log(2)/(log(3/2)) x=log base 3/2 of 2=1.71

Wow, in this sunday morning, I am the first one to watch and listen this bright lady.

When m^3 = 2, take the cube root. So m = cube root 2 or 2^(1/3). Similarly, when m^3 = -1, take the cube root and m = -1

Let m cube be a so the equation becomes a^2 - a - 2 = 0 => a^2 - 2a + a -2 = 0 => a(a-2) + (a-2) = 0 => (a+1)(a-2) = 0 => a = -1 or a = 2 hence, a = m cube = -1 or 2 m = -1 or cube root 2 Ezy

Liked and learnt alot keep up

There's an even cleaner solution: 2^x + 4^x = 8^x 2^x + (2^2)^x = (2^3)^x 2^x + (2^x)^2 = (2^x)^3 Let u = 2^x u + u^2 = u^3 u + u^2 - u^3 = u^3 - u^3 u + u^2 - u^3 = 0 -u(-1 - u + u^2) = 0 -u(u^2 - u - 1) = 0 (-1)(-u)(u^2 - u - 1) = (-1)*0 u(u^2 - u - 1) = 0 u = 0, or u^2 - u - 1 = 0 u = 0 Remember u = 2^x 2^x = 0 No solution 1*u^2 - 1*u - 1 = 0 Let a = 1, b = -1, c = -1 u = (-b +/- sqrt[b^2 - 4*a*c]) / (2*a) u = (-[-1] +/- sqrt[(-1)^2 - 4*1*(-1)]) / (2*1) u = (1 +/- sqrt[1 + 4]) / (2) u = (1 +/- sqrt[5]) / 2 u = (1 + sqrt[5]) / 2, or u = (1 - sqrt[5]) / 2 Remember, u = 2^x 2^x = (1 + sqrt[5]) / 2, or 2^x = (1 - sqrt[5]) / 2 (1 - sqrt[5]) / 2 < 0, so 2^x =/= (1 - sqrt[5]) / 2 2^x = (1 + sqrt[5]) / 2 log(2^x) = log([1 + sqrt(5)] / 2) x*log(2) = log([1 + sqrt(5)] / 2) x*log(2)/log(2) = log([1 + sqrt(5)] / 2) / log(2) x*1 = log([1 + sqrt(5)] / 2) / log(2) x = log_2([1 + sqrt(5)] / 2) x = log_2(1 + sqrt[5]) - log_2(2) x = log_2(1 + sqrt[5]) - 1

Well explained but why "Olympiad" in the title? It's just an elementary exponential equation. Please do not do clickbaits

Great solutions ❤

W^3 + 4W - 39 = 0 (W - 3)(W^2 + 3W + 13) = 0 W = 3, (-3 +/- i √43)/2

Ma'am I am sure you know fuck all about mathematics. Please read the comments

I like the fact that before you split 4W you explained the logic that 13 was a factor of 39.

Of course, instead of creating the quadratic, x^2-6x-40, you had another difference of squares of ((x-3)^2-(7)^2), which could factored into ((x-3)+7)(x-3)-7), which creates the factors (x+4)(x-10), same result.

People keep saying how they can guess some of the answers, which is great, but this is a lesson on how to use the systematic methods of algebra to find the solutions. You might be able to guess on simple ones like this, but a systematic approach teaches how to solve ones you can’t guess.

Great and nice approach. The verification is awesome.

Slightly different method: 2^(3^[5^x]) = 3^(2^[5^x]) log(2^[3^(5^x)]) = log(3^[2^(5^x)]) 3^(5^x) * log(2) = 2^(5^x) * log(3) 3^(5^x) * log(2) / log(2) = 2^(5^x) * log(3) / log(2) 3^(5^x) * 1 = 2^(5^x) * log(3) / log(2) 3^(5^x) = 2^(5^x) * logbase(2)of(3) 3^(5^x) / 2^(5^x) = 2^(5^x) / 2^(5^x) * logbase(2)of(3) 3^(5^x) / 2^(5^x) = 1 * logbase(2)of(3) (3/2)^(5^x) = logbase(2)of(3) log([3/2]^[5^x]) = log(logbase[2]of[3]) 5^x * log(3/2) = log(logbase[2]of[3]) 5^x * log(3/2) / log(3/2) = log(logbase[2]of[3]) / log(3/2) 5^x * 1 = log(logbase[2]of[3]) / log(3/2) 5^x = logbase(3/2)of(logbase[2]of[3]) log(5^x) = log(logbase[3/2]of[logbase(2)of(3)]) x * log(5) = log(logbase[3/2]of[logbase(2)of(3)]) x * log(5) / log(5) = log(logbase[3/2]of[logbase(2)of(3)]) / log(5) x * 1 = log(logbase[3/2]of[logbase(2)of(3)]) / log(5) x = logbase(5)of(logbase[3/2]of[logbase(2)of(3)]) x ~= 0.079164

Correction: x ~= 0.079, not 0.79.

ln1=0

2*x^3 + 2*x = 20 2(x^3 + x) = 2*10 1/2*2*(x^3 + x) = (1/2)*2*10 x^3 + x = 10 x^3 + x - 10 = 10 - 10 x^3 + x - 10 = 0 x^3 + x + (-8 - 2) = 0 x^3 + x + (-2^3 - 2^1) = 0 (x^3 - 2^3) + (x - 2^1) = 0 (x - 2)(x^2 + x*2 + 2^2) + 1*(x - 2) = 0 (x - 2)(x^2 + 2x + 4) + 1*(x - 2) = 0 (x - 2)([x^2 + 2x + 4] + 1) = 0 (x - 2)(x^2 + 2x + [4 + 1]) = 0 (x - 2)(x^2 + 2x + 5) = 0 x - 2 = 0, or 1*x^2 + 2x + 5 = 0 Let a = 1, b = 2, c = 5 x - 2 = 0, or x = (-b +/- sqrt[b^2 - 4*a*c]) / (2*a) x - 2 + 2 = 0 + 2, or x = (-2 +/- sqrt[2^2 - 4*1*5]) / (2*1) x = 2, or x = (-2 +/- sqrt[4 - 20]) / (2) x = 2, or x = (-2 +/- sqrt[-16]) / 2 x = 2, or x = (-2 +/- sqrt[2^4*(-1)]) / 2 x = 2, or x = (-2 +/- sqrt[2^4]*sqrt[-1]) / 2 x = 2, or x = (-2 +/- 2^2*i) / 2 x = 2, or x = -2/2 +/- 2^2*i/2 x = 2, or x = -1 +/- 2*i x = 2, or x = -1 + 2*i, or x = -1 - 2*i x = 2, or x = -(1 - 2i), or x = -(1 + 2i) x1 = 2 x2 = -(1 - 2i) x3 = -(1 + 2i)

4^y = 30 log(4^y) = log(30) y*log(4) = log(30) y*log(4)/log(4) = log(30)/log(4) y*1 = log(30)/log(4) y = log(2*3*5)/log(2^2) y = (log[2] + log[3] + log[5])/(2*log[2]) y = log(2)/(2*log[2]) + log(3)/(2*log[2]) + log(5)/(2*log[2]) y = 1/2*log_2(2) + 1/2*log_2(3) + 1/2*log_2(5) y = 1/2*1 + 1/2*log_2(3) + 1/2*log_2(5) y = (1/2)*(1 + log_2[3] + log_2[5]) y =(1 + log_2[3] + log_2[5]) / 2

Very concise!

Trying to find out Y people failed. 😆

It"s suickly done buy using Hörner method. You should try