How To Solve A Homemade Exponential Equation

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[@FisicTrapella]

b/a = sqr(7) (9:46)

[@mcwulf25]

Yes, not ÷4.

[@MichaelJamesActually]

I agree

[@scottleung9587]

Wow, that was a doozy! Nice job, though.

[@SyberMath]

Thanks!

[@tarunmnair]

I think the value of x is, x=-1/2 +i.arctan(sqrt(7))/ln2 ... that 4 gets cancelled when we do b/a...

[@mcwulf25]

Could see there was no real solution because 2 is not equal to 4. And x=0 doesn't make it go away.

[@SyberMath]

could've had some logarithmic solutions

[@mcwulf25]

@@SyberMath how would that work?

[@mcwulf25]

@SyberMath OK so you ln both sides a few times and get x * (some weird function of ln(2)) = x * (some weird function of ln(4))

[@robertibatullin8329]

Generally speaking, are there any real a and b such as ab, a1, b1, and the equation a^a^a^x = b^b^b^x has a real solution? An intuitive answer is no, but can it be proven?

[@robertibatullin8329]

The answer is yes, I've found one case and there are infinitely many of them. For a=256, b=256^2, the equation a^a^a^x = b^b^b^x has a solution x=-1/4

[@Qermaq]

From the thumbnail: 2^2^2^x = 4^4^4^x appears to be nonsense. If it were 2^x = 4^x we could easily say x = 0. But beyond that, any change we make on one side is blown up on the other. So I'm stumped for now. I think there's no real solutions.
5:10 yep I called it. Have fun! I don't chase after all real adversaries!

[@juergengeisslinger5051]

HELLO
ARE YOU SURE, THAT THE ANGLE PHI IS CORRECT?

[@SyberMath]

which angle?

[@denisseveliz2007]

Hello, my boyfriend really like your channel and spend a lot of time solving your problems. His birthday will be soon, asking you our the community to create a problem and I can decorate his birthday cake with it. I hope you can guys help me. Thank you 🫶

[@cav1928]

I made it with natural logarithms, and applied it several times. It is ease for me to think in log terms that in tower exponents, at the end of course I need to do a variable change and the result is the same. Sincerely at first glance I guess that my solution was wrong because it was very long for a problem that seems at first sight simple.

[@giuseppemalaguti435]

Posto t=2^x, risulta t=1+2t^2 che pero non da soluzioni reali...boh

[@theCzechoslovak]

Try to guess the solution
The solution:

[@SyberMath]

😮😁

[@yoav613]

Nice,but i hope you do not ask problems like this in your tests😂

[@Qermaq]

I have a problem with the inverse tangent being written as tan(-1)theta. If we want the tan of an angle squared, it's tan(a^2). If we want the tan of a to be squared, it's tan^2(a). So if I want the tan to be raised to the power -1, I must write tan^(-1)(a). So it's ambiguous.

[@richardguimond7665]

Indeed I prefer the inverse tangent being written as arctan because there is no risk of ambiguity.

[@Palkia8-Bit]

I am still convinced you’re tigerofwind XD

[@SyberMath]

😁😉

[@penguin9257]

2=4

[@monasimp87]

Really nice solution

[@StaR-uw3dc]

Nice.
Slightly changed equation: 2^(4^(2^x)) = 4^(2^(4^x)) has the real root.

[@SyberMath]

Thank you!

[@SyberMath]

Here comes the video. Thank you for the suggestion...❤️😍
Will go public in less than 8 hours:
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[@StaR-uw3dc]

@@SyberMath Nice to hear it.

[@adithyan9263]

Cool