Solving An Infinite Exponential Equation

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[@SuperMtheory]

looks like Lambert W

[@senpai7276]

I think it's when x^x = y

[@chaosredefined3834]

y = 2^(x^y) ln y = x^y ln 2 ln ln y = y ln x + ln ln 2 ln ln y - ln ln 2 = y ln x ln (ln y/ln 2) = y ln x (ln (ln y/ln 2))/y = ln x (ln y/ln 2)^(1/y) = x (log2 y)^(1/y) = x We want y = 4. So... (log2 4)^1/4 = x 2^1/4 = x This also gives you a mechanism to calculate any equation of this form, just changing the value of y.

[@alinayfeh4961]

Yeah nice a solving an infinite experation equation 😊 , Mild and easy 👌 (paradise to God)

[@siguardvolsung]

This video raises more questions than it solves.

[@Qermaq]

Re: last question, I don't think "infinitely many". Seems to me as the curve goes up at the right it will converge to 1. Consider any curve similar to 2^a-x, where a is any real value more than one. x values between 1 and 2^0.25 spit out two y values. Just below 2^0.25 we get y = 4-d or 4+d. As we approach one from the right, 4-d approaches 2 and 4+d approaches infinity. At y=1 the upper option has diverged and it's just x=2.

[@felaxwindwalker206]

Enjoyed the video. So you know, you actually can set the entire exponent equal to 2, leaving you with x^2^2=2, which simplifies to x^4=2, or x=2^(1/4). The exponent in this case remains an infinite pattern, you've just reversed the order of the pattern.

[@felaxwindwalker206]

Oh, and the negative equivalent, as well as two complex solutions.

[@popitripodi573]

Very nice ❤an easy one but food for thought

[@Schaex1]

I also found ±2^(0.25) i as complex solutions to this problem. Is this valid?

[@giuseppemalaguti435]

X=(1/4)log(2)4

[@mathswan1607]

x= +/- fourth root of 2

[@barakathaider6333]

👍