Math Olympiad | A Nice Algebra Problem | A Nice Radical Problem

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Find the value of x?
How to solve x^√x=√x^x

In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation x^√x=√x^x
in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #math #maths

[bestseens]

I have a very great solution , please react on it!

Given, x^(√x) = √(x^x)

In the R.H.S square root is taken within a number with power,

By using property,

Therefore, √(x^x) can be written as x^(x-2)

x^(√x)= x^(x-2)
Here base is same,

Therefore, powers are also same

√x = x-2
Squaring on both sides

x = x²-4x+4
x²-5x+4=0
Solving quadratic equation,
Gets, (x-1)(x-4) =0
By distributing,

x = 1 and x= 4

[A_Weird_Blue_Guy]

x^√x = √(x^x)
x^√x = x^x/2
√x = x/2
x = x²/4
x/x = (x²/4)/x
1 = x²/4x
1 = x/4
4 = x
x = 4

U could also get x = 0 by simply looking at the equation

[stevenmayhew]

When you raise a number to the zero power, what you are really doing is dividing that number by itself, which is where the 1 comes from. However, zero to the zero power is zero divided by zero, which is considered to be indeterminate. I see zero divided by zero to be all numbers. So, zero to the zero power is also all numbers.

[quixata]

What nonsense.... who told you 0^0 = 1? This is purely absurd as 0^0 is undefined. x=0 is not a solution to the given equation. 0 is outside the domain. Rather x=1 is indeed a solution besides x=4.

[MariaJuanaLinares]

¡Cero elevado a la cero no tiene sentido! No está definido en los números reales

[ashoq]

isn't 0 to the power 0 undefined ? Or is that only valid for limiting vales ?

[otabekpardabaev]

x=1 doesn’t have to come from inspection.
x^sqrt(x)=x^(x/2)
Ln (x^sqrt(x))=ln(x^(x/2))
sqrt(x)*Ln (x)= (x/2)*ln(x)
Ln(x)*(sqrt(x)-(x/2))=0
Ln(x)=0, x1=1.
The remaining is as what you did.

[key_board_x]

x^(√x) = √x^(x)
x^[x^(1/2)] = [x^(x)]^(1/2)
x^[x^(1/2)] = x^(x/2)
x^(1/2) = x/2
[x^(1/2)]^(2) = (x/2)^(2)
x = x²/4
x² = 4x
x² - 4x = 0
x.(x - 4) = 0

First case: x = 0
Second case: (x - 4) = 0 → x = 4

[Master_Unknown]

What type of teacher u are ? How can 0^0=1 ? Don't u know anything about indeterminate forms ? And, if u don't know about it, then why are u making this useless video ?