Hope your pet cat is doing fine.

This can also be solved using the Lambert W function. First take ln of both sides. Thus x*ln(x)=(1/x)*ln(2) or (x*2)*ln(x) = ln(2). Now let t=x^2. That gives t*ln(t^(1/2)) = ln(2), Simplifying gives t*ln(t) = 2*ln(2). This equation can be written as ln(t)* e^ln(t) = (ln(2))*e^ln(2). If we take the lambert W of both sides W( ln(t)* e^ln(t)) = W( (ln(2))*e^ln(2)) or ln(t)= ln(2) or t=2 or x^2 = 2 or x=sqrt(2). The advantage of this method is the argument of W( (ln(2))*e^ln(2)) is in the branch of the W function where there is only a single answer. Thus x=sqrt(2) is the only answer and the calculus approach can be avoided.

@YouTube_username_not_found

2 күн бұрын

Actually, you need calculus to find out how many solutions in each branch of W function.

@HarisRehmanGG

Күн бұрын

You didn't tell us the bprp fish introduction before using it 😭

Made me wonder if there's a straight line that crosses x^x^2 in three places and yes, there are infinitely many. One simple example is: y = -x/3 + 1.01 If 0^0^2 could be said to be exactly 1, and that was then allowed to count as a crossing point, then the 1.01 could just be 1, but that would only anger the mathematical gods. And many fellow KZread commenters. As for finding those crossing points algebraically, I don't think even WolframAlpha can help here. Except if it does so numerically.

😊😊😊👍👍👍

x^x = 2^(1/x) x^x² = 2 x²^x² = 2² x² = 2 => *x = √2* (x = -√2 is not valid)

@xualain3129

3 күн бұрын

Bravo! I got exactly the same idea as yours.

@mcwulf25

2 күн бұрын

This is how I solved it 👍

@Anmol_Sinha

2 күн бұрын

why is x = -√2 not valid?

@mcwulf25

Күн бұрын

@Anmol_Sinha when both sides were squared, that extraneous solution was added. The rhs is always positive. The lhs must also therefore be positive.

@SidneiMV

Күн бұрын

@@mcwulf25 exactly. tks

x^x=2^(1/x) Raise to x --> x^x²=2 y=x² --> x=y^(½) [y^(½)]^y=2 y^(½y)=2 ½y^(½y)=1 --> ½y=1 y=2 x²=2 --> x=±sqrt(2)