วีดีโอ

i can't believe its 2φ!

An insult to math.

I fail to see any value in this "solution".

If you are looking only for real-valued solutions, only the one with 1+sqrt(5) is valid. Might be worth mentioning.

Isn‘t it just X=3

Important: for step 1 you need to test x=0 separately, slamming a division by x into your solution can cause problems down the line. Luckily 0^0 is undefined so it doesn’t matter here

yeah... why don't we just say it's -3

I took the 4th root of X^4 on the left side and the 4th root of 1 in the right side and got X=1.

x^x = x^2 therefore x = 2 Solved in 3 seconds

i did it on my head before clicking the video, just wanted to confirm, cool problem!

How I approached this problem: Check that for x>0 the value is too small. Check for x<0 the values x=-2 and x=-3. See that x=-3 solves the equation. Very hard problem^^

x = -3

It's 11108889, you forgot to add the last three digits of the product of 333x333.

got it

I had been spending 15 mins for solving before noticed that title of video has x^2-x^3=100 instead of x^3-x^2🙂

You added 40400 + 760 wrong it should be 41160

That's pretty complex.

8^x=x^6 cbrt all sides 2^x=x^2 x=2 is a valid solution to this, but lets solve algebraically. ln all sides xln2=2lnx divide by x on all sides ln2=2lnx / x divide by 2 on all sides (ln2)/2=(1/x)lnx raise e to both sides sqrt2=x^(1/x) ok so for questions along the format “x^1/x=y” its a long proof, but the answer is always x=e^-W(-ln2) so x=e^-W(-lnsqrt2)=e^-W(-0.5ln2) which, simply, =2.

how about using los?? seems the way to go without using a calculator for approximate answers

C'mon man - pull out the old LambertW. This is rubbish. I feel like I lost a few brain cells watching it.

I got X^18= 9, take the 18th root of 9 on each side and X = 1.12983. To check, the 18th root of 9 = 1.12983 raised to 1.12983 then raised to 18 = 3.

Math classes will never accept an approximation unless it's is specifically asked for.

This is incomplete as you miss solutions. The technical way to do it is: xln(8) = 6ln(x) = ln(8)exp(ln(x)) ln(8) = 6ln(x)exp(-ln(x)) -ln(8)/6 = -ln(x)exp(-ln(x)) ==> W(-ln(8)/6) = -ln(x) ==> x = exp(-W(-ln(8)/6)) Using both real branches W_-1 and W_0 of Lambert-W should give u other real solutions such as x = 4 (edit: I think using a complex branch of the Lambert-W may also give a negative real solution for x too)

2

What has this to do with math olympiad? They really ask such easy stuff there?

Quick question: What would make anyone think of doing the equation that way? why would anyone think of simplifying 8 to (2^3)? Why wouldn't anyone with an arithmetic background do it this way instead: 8^x=x^6 =>>(8^x)/(x^6)=0 =>>> + - x=[(8^x)/(x^6)]; and end the equation there and be correct?

You left out two other solutions: x = 4 and x ≈ -0.766665. All three solutions, with an exact representation of that last, negative solution, can be found using the Lambert W function.

damn bro this was rlly helpful and rlly fun to watch dont give up uploading surely u will blow up and also i always watch ur videos their pretty sick

Nice, but not very formal. In order to be sure that 16^(1/16) is the only solution, you'd also have to prove that the function f(x)=x^x^16 is injective. Otherwise, you have no way of knowing whether or not there are any other real solutions.

I think you're missing an important step. You should have said, if a^a = b^b *and the b in question is bigger than 1,* then a=b. Because the function x^x is not completely one-to-one. But its subset for x>1 *is* one-to-one.

Yes I got cube root of 3. Very much enjoyed the video.

to resolve algebra problem it’s better to know some algebra first

X = 3,597285023540419

I’d like to see a video on how to verify the solution to the initial problem.

want more such questions

2^(2^2) = 16 16 = x^(x^16) x = 2^(.25) checks out now that I think about it, lambert applies here too. Want e^u * u on the right so we can just get u. where u is in terms of x. ln(16) = x^16 * ln x ln(16) = e^(ln[x^16])*lnx; Mult each side by 16 16ln(16) = e^(ln[x^16]) * ln[x^16] (True by log props, which is why we multiplied by 16) W(ln16^16) = ln(x^16) e^[W(ln16^16)] = x^16 x = (e^[W(16ln16)])^1/16 = (ln(16^16)/[W(ln(16^16))])^1/16 # e^(W(x)) = x/(W(x)) Yeah i think lambo is defined like it is so that cant really be evaluated from there without wolfram

Right-click. Never recommend chanel again/

You advertise a math video and then it's just a bunch of rounding. Yawn

= cube root of 3

log3/log2 = ~1.5

Or 7th root of two, right? The same negative too

You didn't teach us how to solve the problem. You didn't use algebra. And your answer was wrong.

hahaha. i guess there are all kinds of videos out there :)

You give no indication of how you "magicked" the 3.274. These problems are usually solved by using the Lambert W function : W(ψe^ψ)=ψ so we look to construct a suitable ψ x=4^(x/5)=e^(x ln(4) /5) divide both sides by the e value xe^-(x ln(4) / 5) = 1 force the coeff. of the exponent and the x value to be identical by multiplying by -ln(4) /5 -x ln(4) /5 e^-(x ln(4) / 5) = -ln(4) /5 apply the W function (algebraically to the lhs, numerically to the rhs - see Wolfram alpha) -x ln(4)/5 = -0.423425164316885734 x = 0.423425164316885734 (5/ln(4)) = 1.5271834618689137

I solved the equation the same way, but I think it's important to make restriction on exponent-power function (x>0). So if you got +-sqrt8(4) or +-sqrt4(2), negative values won't be the answer.

Cue the slide whistle!

I did it in my mind in 2 seconds

X ≃ 3.5972850235404175

Agree with the two previous comments. 2:40 you cannot simply split the denominator (log 2 + log3) into two denominators If you are going to take logs anyways - why not simply log12/log6?

Go back to school and take up basket weaving.