Solving x^{4x^8}=2 | Math Competitions

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[IorPerry]

to be precise the solution are 8: 2^(1/8)*e^(n*i*pi/4) with n between 0 and 7 😁

[protoman]

In my head, I noticed that if we do x = 8th root of 2, then x^8 * 4 = 8, and thus you have x^8 = 2, which fits.

Of course, one thing this channel has taught me is to confirm all solutions instead of 1, so naturally there are 8 actual 8th roots of 2. So thus the solution set should be all 8 such roots, and they can be found using complex numbers.

[gelbkehlchen]

Solution:
x^(4*x^8) = 2 |()² ⟹
[x^(4*x^8)]² = 2² ⟹
x^(2*4*x^8) = 2² ⟹
x^(8*x^8) = 2² ⟹
x^8^(x^8) = 2² |The same operations are done with x^8 on the left side of the equation as with 2 on the right side of the equation, therefore is ⟹
x^8 = 2 |()^(1/8) ⟹
x1/2 = ±2^(1/8)

[Anmol_Sinha]

The generalization was a great addition to this video.

[nasrullahhusnan]

In 2:59 don't you think it's simpler if we let y=x⁸ giving y^y=2^2 and hence y=2 --> x⁸=2 --> x=2^⅛?

[seba]

this time i m going to resolve it without watching i have seen a lot of your math videos great channel

[popitripodi]

Very nice once again!!! Exponential equation and calculus!!!!❤❤❤❤

[jeremygalvagni]

Note: it's nice you didn't show a Desmos graph, since it doesn't show the negative solution. (negative)^(x) isn't real for many values of x.

[iRReligious]

If you have 2 'x' at end as a solution, why only 1 intercept point on the graph?! 📈🤔🧐🤯

[mbmillermo]

But if we have an 8th degree polynomial, aren't there 8 solutions? Probably @IorPerry has it right.

[mihaleben]

Woah, wait... i just, found how to disprove it?
1 wouldnt work
2 would be a too big number
And a decimal number would get you a smaller number.
So, no, it is not possible