Understanding the Exponential Equation

We are asked to find the set of solutions for the equation \( {{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} \), where n is a positive integer. This is an exponential equation involving roots.

Simplifying the Equation

First, let's simplify the right side of the equation. The nth root can be written as a fractional exponent:

\( \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} = {{\left( {{{\rm{x}}^{\rm{x}}}} \right)}^{1/{\rm{n}}}} \)

Using the power of a power rule \( {(a^b)^c = a^{bc}} \), we get:

\( {{\left( {{{\rm{x}}^{\rm{x}}}} \right)}^{1/{\rm{n}}}} = {{\rm{x}}^{{\rm{x}} \cdot (1/{\rm{n}})}} = {{\rm{x}}^{{\rm{x}}/{\rm{n}}}} \)

So, the original equation becomes:

\( {{\rm{x}}^{\sqrt {\rm{x}} }} = {{\rm{x}}^{{\rm{x}}/{\rm{n}}}} \)

Solving the Simplified Equation

For an equation of the form \( a^b = a^c \), there are typically a few possibilities for the solutions:

• The base is 1: \( a=1 \). In this case, \( 1^b = 1^c \) is always true for any valid exponents b and c.
• The exponents are equal: \( b=c \), provided the base \( a \) is not 0, 1, or -1 (where issues with exponents might arise).
• The base is -1: \( a=-1 \), and both exponents \( b \) and \( c \) are either both even or both odd.
• The base is 0: \( a=0 \), and both exponents \( b \) and \( c \) are positive. \( 0^0 \) is usually undefined.

In our equation \( {{\rm{x}}^{\sqrt {\rm{x}} }} = {{\rm{x}}^{{\rm{x}}/{\rm{n}}}} \), the base is \( x \). The exponents are \( \sqrt{x} \) and \( x/n \).

Since \( \sqrt{x} \) is in the exponent, we must have \( x \ge 0 \). This eliminates the case where the base is -1.

Case 1: The base \( x \) is 1

If \( x=1 \), the equation becomes:

\( {{1}^{\sqrt {1} }} = {{1}^{1/{\rm{n}}}} \)

\( {{1}^{1}} = {{1}^{1/{\rm{n}}}} \)

\( 1 = 1 \)

This is true. So, \( x=1 \) is a solution.

Case 2: The exponents are equal

If the base \( x \) is not 0 or 1, we can equate the exponents:

\( \sqrt{\rm{x}} = {\rm{x}}/{\rm{n}} \)

To solve for \( x \), we can square both sides (remembering that squaring can introduce extraneous solutions, so we'll need to check our answers):

\( {(\sqrt{\rm{x}})}^2 = {({\rm{x}}/{\rm{n}})}^2 \)

\( {\rm{x}} = {{\rm{x}}^2}/{{\rm{n}}^2} \)

Now, rearrange the equation to solve for \( x \):

\( 0 = {{\rm{x}}^2}/{{\rm{n}}^2} - {\rm{x}} \)

Factor out \( x \):

\( 0 = {\rm{x}}\left( {{\rm{x}}/{{\rm{n}}^2} - 1} \right) \)

This gives two possibilities for \( x \):

• \( {\rm{x}} = 0 \)
• \( {{\rm{x}}/{{\rm{n}}^2} - 1} = 0 \Rightarrow {{\rm{x}}/{{\rm{n}}^2}} = 1 \Rightarrow {\rm{x}} = {{\rm{n}}^2} \)

Let's consider \( x=0 \). The original equation is \( {{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} \). If \( x=0 \), we get \( {{0}^{\sqrt {0} }} = \sqrt[{\rm{n}}]{{{{0}^{0}}}} \). Both sides involve \( 0^0 \), which is typically considered an indeterminate form and undefined in this context. For \( x^{\sqrt{x}} \) to be defined in general for \( x \ge 0 \), we usually consider \( x > 0 \) for \( x^{\sqrt{x}} \) unless \( x \) is an integer and the exponent is positive. Since the exponent \(\sqrt{x}\) involves a root, \(x=0\) is not usually considered a valid solution stemming from equating exponents in this form. The condition \( x > 0 \) or \( x=1 \) is generally required for \( x^a = x^b \implies a=b \).

So, we focus on the other solution from this case: \( x = n^2 \).

Verifying the Solution \( x = n^2 \)

Substitute \( x = n^2 \) back into the original equation \( {{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} \). Remember n is a positive integer.

Left Hand Side (LHS): \( {{\rm{x}}^{\sqrt {\rm{x}} }} \)

Substitute \( x=n^2 \): \( {{({{\rm{n}}^2})}^{\sqrt {{{\rm{n}}^2}} }} \)

Since n is a positive integer, \( \sqrt {{{\rm{n}}^2}} = {\rm{n}} \).

LHS \( = {{({{\rm{n}}^2})}^{\rm{n}}} = {{{\rm{n}}^{2 \cdot {\rm{n}}}}} = {{{\rm{n}}^{2{\rm{n}}}}} \)

Right Hand Side (RHS): \( \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} \)

Substitute \( x=n^2 \): \( \sqrt[{\rm{n}}]{{{{({{\rm{n}}^2})}^{{{\rm{n}}^2}}}}} \)

\( = \sqrt[{\rm{n}}]{{{{\rm{n}}^{2 \cdot {{\rm{n}}^2}}}}} = \sqrt[{\rm{n}}]{{{{\rm{n}}^{2{{\rm{n}}^2}}}}} \)

Convert the nth root to a power:

\( = {{\left( {{{\rm{n}}^{2{{\rm{n}}^2}}}} \right)}^{1/{\rm{n}}}} = {{{\rm{n}}^{{(2{{\rm{n}}^2})/{\rm{n}}}}}} = {{{\rm{n}}^{2{\rm{n}}}}} \)

Since LHS = RHS (\( {{{\rm{n}}^{2{\rm{n}}}}} = {{{\rm{n}}^{2{\rm{n}}}}} \)), \( x = n^2 \) is a solution.

Set of Solutions

We found two valid solutions for the equation: \( x=1 \) and \( x=n^2 \).

The set of solutions is \( \{1, n^2\} \).

Revision Table: Key Concepts in Solving Exponential Equations

Additional Information: Domain of Exponential and Radical Expressions

When dealing with expressions like \( x^y \) and \( \sqrt[n]{x} \), it's important to consider the valid values for the base \( x \) and the exponents/roots.

• For \( \sqrt{x} \), the term is defined for \( x \ge 0 \) in real numbers.
• For \( x^y \):
  • If \( y \) is an integer, \( x^y \) is defined for most \( x \). If \( y < 0 \), \( x \neq 0 \).
  • If \( y \) is a rational number \( p/q \) (in lowest terms), \( x^{p/q} = \sqrt[q]{x^p} \). If \( q \) is even, we usually need \( x \ge 0 \) for real values. If \( q \) is odd, \( x \) can be any real number.
  • If \( y \) is an irrational number, \( x^y \) is typically defined only for \( x > 0 \).

In our problem, the equation \( {{\rm{x}}^{\sqrt {\rm{x}} }} = \sqrt[{\rm{n}}]{{{{\rm{x}}^{\rm{x}}}}} \) contains \( \sqrt{x} \), which requires \( x \ge 0 \). It also contains \( x^{\sqrt{x}} \), where the exponent \( \sqrt{x} \) can be irrational (e.g., if x=2). This reinforces the constraint \( x > 0 \) for this form, although \( x=1 \) is a special case covered separately. The term \( x^x \) also appears, which is usually considered for \( x > 0 \) or \( x=1 \) (where \( 1^1=1 \)). The nth root \(\sqrt[n]{x^x}\) is defined for \(x^x \ge 0\) if n is even, which implies \(x \ge 0\) if \(x^x\) is defined for \(x \ge 0\). If n is odd, it's defined for all \(x\) for which \(x^x\) is defined. Given \(n\) is a positive integer, the \( \sqrt{x} \) term is the strongest constraint, requiring \( x \ge 0 \). Our solutions \(x=1\) and \(x=n^2\) satisfy \(x \ge 0\) (since n is positive). \(x=n^2\) will be strictly positive if \(n \neq 0\), which is true as n is a positive integer. \(x=1\) is also positive.