A Nice Exponential Equation

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[allanmarder]

If you let y=sqrt(x) the equation becomes (y^2)^(1/y) =2. then raising both sides to the power of y, you get y^2 = 2^y and SyberMath actually presented this problem about a year ago.
(I don't know the exact date). The solutions were y=2 and y=4. In this case the x=y^2 so that x=4 and x=16.

[AdityaKumar-iypu]

By analysing the question i.e put different values to check it, i got answer within 15 seconds 😌. X = 4

[ahmetleventtakr]

That Turkish name pronounciation was flawless, any chance you are from Turkey?

[TheMoogleKing]

There's a trick that comes up on a lot of these videos that works here. Take both sides to the power of - 1/2, then let a = 1/ root x and you have a ^ a = (1/2)^(1/2), so 1/root(x) = (1/2) and x = 4.

[MichaelRothwell]

My immediate thought was to convert the equation x^(1/√x)=2 into an equation of the form yʸ=a, as we would then be dealing with a much more familiar function.
So let y=1/√x, y²=1/x, x=1/y²
where x>0 and y>0.
Substituting in the original equation,
(1/y²)ʸ=2
1/y²ʸ=2
y²ʸ=½
yʸ=½^½
Clearly y=½ is a solution.
Also we know (using calculus) that the function xˣ is strictly decreasing from the value 1 at x=0 down to x=1/e then strictly increasing for x>1/e, passing through the value 1 at x=1.
So for x>0 there are exactly two solutions to xˣ=a for all values of of xˣ<1 except for when x=1/e.
Since ½^½<1 and ½≠1/e, there are exactly two solutions to yʸ=½^½.
Again, by familiarity with the function xˣ we know that the other solution is y=¼, as ¼^¼=(½²)¼=½^½.
Since x=1/y², there are two solutions to the given equation and they are x=4 & x=16.

[joyli]

3:16

In this time of the video, we get ln(x)/sqrt(x) = ln(4)/2, which that we can find out that in the upper part; we can directly find out that x is equal to 4, and checking the answer 2^2 = 4 is correct. So x is equal to 4.

[walterufsc]

Here is my solution (without using logarithms):
x^(1/√x) = 2
Raising both sides to power √x:
x = 2^√x
By making y = √x ==> x = y²
So y² = 2^y (i)
From (i), we immediately have: y = 2 ==> x = 4
Squaring (i): (y²)² = (2^y)² ==> y^4 = 2^2y = 4^y (ii)
From (ii), we immediately have: y = 4 ==> x = 16

[zelda]

So obviously f(a) = f(b) -> a = b IFF f(x) is a bijection. My question is how do you determine if a composition of functions is a bijection or not without actually testing some salient points?

[wastedontheyoung]

I enjoyed other videos you produced where you simplified the equation before solving it.
You arrived at similar answers in shorter videos without using logarithms.
Would this be a more practical strategy?

[carlyet]

my thought on inspection was log base two
log_2 x = x^[1/2] graph LHS & RHS and consider the relations

[ivormacky]

All that you need in this problem is to take derivative from x^1/sqrt(x)

[joyli]

According to graphing the function, we can find out that variable x is equal to 4.

[hxd]

what program do you use for the board?