If $x^y = y^x$, then $$ \left(\frac{x}{y}\right)^{\frac{x}{y}} $$ is equal to

### Concept & Formula This problem tests the manipulation of fractional bases and exponents. The primary strategy is to use the given relation $x^y = y^x$ to substitute the denominator $y$ in the target expression with a term strictly based on $x$, thereby simplifying the expression to a single base. ### Step-by-Step Solution * **Step 1:** Expand the target expression by applying the exponent to both the numerator and the denominator. $$ \left(\frac{x}{y}\right)^{\frac{x}{y}} = \frac{x^{\frac{x}{y}}}{y^{\frac{x}{y}}} $$ * **Step 2:** Use the given equation $x^y = y^x$ to find an expression for $y$ in terms of $x$. Take the $1/x$ power of both sides: $$ (y^x)^{\frac{1}{x}} = (x^y)^{\frac{1}{x}} $$ $$ y = x^{\frac{y}{x}} $$ * **Step 3:** Substitute this new expression for $y$ into the denominator of our expanded target expression. Denominator: $ y^{\frac{x}{y}} = \left(x^{\frac{y}{x}}\right)^{\frac{x}{y}} $ * **Step 4:** Simplify the new denominator using the power rule $(a^m)^n = a^{mn}$. Multiply the exponents: $\frac{y}{x} \cdot \frac{x}{y} = 1$. So, the denominator simply becomes $x^1 = x$. * **Step 5:** Reassemble the fraction and simplify using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$. $$ \frac{x^{\frac{x}{y}}}{x^1} = x^{\frac{x}{y} - 1} $$ ### Exam Strategy & Shortcut **Smart Value Substitution:** Pick numbers that satisfy $x^y = y^x$. The classic non-trivial integer pair is $x=4$ and $y=2$ (since $4^2 = 16$ and $2^4 = 16$). Plug these into the target expression: $$ \left(\frac{4}{2}\right)^{\frac{4}{2}} = (2)^2 = 4 $$ Now plug $x=4$ and $y=2$ into the options to see which yields 4: (a) $4^{2/4} = 4^{1/2} = 2$ (Incorrect) (b) $4^{(4/2) - 1} = 4^{2-1} = 4^1 = 4$ (Correct!) (c) $1$ (Incorrect) (d) $4^{4/2} = 4^2 = 16$ (Incorrect) This method is incredibly fast and completely avoids tricky algebraic exponent tracking. ### Common Pitfall Students often get tangled in the exponents during Step 3, misapplying the $(a^m)^n$ rule and failing to realize that multiplying the reciprocal exponents $\frac{y}{x}$ and $\frac{x}{y}$ perfectly cancels them out to $1$. ### Final Answer Therefore, the correct answer is **$x^{\frac{x}{y}-1}$**.