Solving a Radical Exponential Equation

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #Radicals #RadicalEquations #ExponentialEquations
via @YouTube @Apple @Desmos @NotabilityApp
@googledocs @canva

PLAYLISTS 🎵 :

[GirishManjunathMusic]

Before watching the video:
Given:
√(x↑√x) = x↑(1/√x)
To find:
x
Flattening exponents:
x↑(√x/2) = x↑(1/√x)

Taking natural log on both sides:
(√x/2)·(lnx) = (1/√x)·(lnx)

Moving all terms to LHS:
(√x/2)·(lnx) – (1/√x)·(lnx) = 0
(lnx)(√x/2 – 1/√x) = 0
lnx = 0 or √x/2 = 1/√x
lnx = 0 → x = 1

Or:
√x/2 = 1/√x
As x ≠ 0 for RHS to be defined, Multiplying both sides by √x

x/2 = 1
x = 2.

[tbg-brawlstars]

By seeing it for 10 seconds, I think 1 and 2 are solutions to this....Idk about other solutions

[misterdubity]

Go to Desmos and graph the LHS and the RHS. Two intersections at (1, 1) and (2, ~1.6325).

[pwmiles]

sqrt(x^(sqrt(x)) = x^(1/sqrt(x))

Square both sides
x^(sqrt(x)) = x^(2/sqrt(x))

x=1 is a solution as 1 to power of anything is 1. Otherwise equate the exponents

sqrt(x) = 2/sqrt(x)

Multiply both sides by sqrt(x)
x = 2

[filimongraziano]

For second method from x^x=x² just do ln on both sides s.t. x·ln(x)=2·ln(x) so (x-2)ln(x)=0 therefore x-2=0 V ln(x)=0 => x=2 V x=1

[protoman]

So I see that we can do some notation conversions to make things easier. Root(x^root(x)) can be rewritten as x^(1/2*x^1/2), and x^(1/root(x)) can be rewritten as x^(x^-1/2).

Automatically, we can see that if we want to compare equality, the three edge cases are 1, 0, and -1. Of these three cases, only 1 works. If you have x^a = x^b you have to check these three cases before setting a = b, because x = 1 makes this always true; x = 0 makes it true for both a and b greater than 0; x = -1 makes it true for a and b having the same parity mod 2.

Ignoring those three cases, if x^a = x^b then a = b. Thus, we can set 1/2*x^1/2 = x^-1/2. If we multiply both sides by x^1/2, we get 1/2*x = 1. Finally, multiply both sides by 2 to get x = 2, no other solutions.

The only solutions are therefore 1 and 2.

[crazycat]

Just raising both sides to the 2√x power, x^x=x²

[kanguru_]

Let x=a^2, so a^a=a^(2/a), so a=2/a, as long as a is not 1, and a=sqrt(2) and x=2. x=1 is also clearly a solution

[GodbornNoven]

I squared both sides and solved for x
X=2

[hanwentian]

I found the solutions in 5 minutes and I am a social science student

[tomotsun]

x^x=x^2
x=1 is one of the answer.
If x≠1,
xlogx=2logx
x=2

[myelectronicsworld]

Does solving the equation by taking the log of both sides restrict the solutions? I get also "2" but not "1".

[chandrashekharmehta]

Ig 2 is the ans. I dud in my mind .
Lets check

[hans-rudigerdrzimmermann]

Greetings from me. I love your channel !
I solved a former problem you posted and would like to send the solution. Do you have email?