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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

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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.

Fail

How to solve a problem when you know the answer 😂

What? 2.631 is incorrect. The answer to "x" is 1.387. Next time please check your answer before posting.

If you take 6^2.631 you get approximately 112.

Is there any way to show that 4 is another solution to this problem, except by trial and error? Thanks, a math nerd.

Very approximative...not satiating...

I think you forgot that x also = -⁸√4

4th root of 2, lol...lucky (well, perhaps somewhat educated) guess...

This is not right. 3.16^2 is 10?!? BS. Bad. Rubbish.

Algebra problem, but not an algebra solution. ☹️

Rubbish...

thats helpful

I was overthinking it and trying to go into the 8 complex solutions, but yeah apart from that I reached this answer.

Why not show us how to use Excel for that? That would be far more accurate and useful. The answer is approximately 3,5972850236, and the result of the operation is quite close to 100. But still, all this has nothing to do with the analytical approach.

This is not a solution to the problem. You just “reverse engineered” a known answer. Even though, it is inaccurate. Your solution squared is 99.02794833582002; not 100.

mathematically inaccurate

This is some mathematical thinking I don't think I've ever encountered. It's cool to see the thought process but I also think it is weird that you have to kind of reverse-simplify in order to get the answer.