Problem Statement: If X power y y power z z power X= X power X then find values of X,y,z?

Solution:

We are given that,

X power y y power z z power X= X power X

Let's simplify this expression by taking the logarithm of both sides,

log(X power y y power z z power X) = log(X power X)

Using the logarithmic property, we can simplify this expression further,

y*log(X) + z*log(y) + X*log(z) = X*log(X)

Now, we need to find the values of X, y, and z that satisfy this equation.

Case 1: X=1

If X=1, then the equation becomes,

y*log(1) + z*log(y) + 1*log(z) = 0

This equation has infinitely many solutions for y and z. For example, y=2 and z=1/2 satisfies the equation.

Case 2: X>1

If X>1, then the equation becomes,

y*log(X) + z*log(y) + X*log(z) = X*log(X)

If we assume that y, z > 1, then we can use the AM-GM inequality to get,

y*log(X) + z*log(y) + X*log(z) >= 3*X^(1/3)*log(X) > X*log(X)

This contradicts the given equation, so we must have at least one of y or z <=>

Case 2.1: y=1

If y=1, then the equation becomes,

z*log(1) + X*log(z) = X*log(X)

This simplifies to,

X*log(z)/(X*log(X) - log(z)) = 1

Let's define a new variable, t=log(z), then we can rewrite the equation as,

X*t/(X*log(X) - t) = 1

Solving for X, we get,

X = t/(log(t) - log(log(t)))

This equation has solutions for t > e, where e is the base of the natural logarithm. For example, t=2.5 satisfies the equation, which gives us X=5.29.

Case 2.2: z=1

If z=1, then the equation becomes,

y*log(X) + X*log(1) = X*log(X)

This simplifies to,

y*log(X) = X*log(X)

Since X>1, we can cancel out log(X) from both sides to get y=X. Therefore, any solution of the form (X,X,X) satisfies the equation.

Conclusion:

Therefore, the solutions to the equation are:

- (X,X,X), where X>1
- (1,y,z), where y and z can be any real numbers

Verification:

Let's verify that the solutions actually satisfy the equation.

- (X,X,X):

X^X * X^X * X^X = X^(3X) = X^(X*3) = X^X * X^X * X^X

- (1,y,z):

1^y * y^z * z^1 = y^z = 1