If `x^(mx^(mx^(mx..."to"oo)))=y^(ny^(ny^(ny..."to"oo)))," then "(dy)/(dx)=`

Options

A. `(y)/(x)`
B. `(x)/(y)`
C. `(my)/(nx)`
D. `(ny)/(mx)`

Answer

To solve the problem, we start with the given equation:

\[ x^{mx^{mx^{mx \ldots}}} = y^{ny^{ny^{ny \ldots}}} \]

1. **Define the Infinite Powers**:
Let \( t = x^{mx^{mx^{mx \ldots}}} \) and \( t = y^{ny^{ny^{ny \ldots}}} \). Since both expressions equal \( t \), we can set them equal to each other:
\[ t = x^{mt} \quad \text{and} \quad t = y^{nt} \]

2. **Set Up the Equations**:
From the first equation, we have:
\[ t = x^{mt} \]
From the second equation, we have:
\[ t = y^{nt} \]

3. **Equate the Two Expressions**:
Setting the two expressions for \( t \) equal gives:
\[ x^{mt} = y^{nt} \]

4. **Take the Logarithm**:
Taking the logarithm of both sides:
\[ \log(x^{mt}) = \log(y^{nt}) \]
Using the property of logarithms, this simplifies to:
\[ mt \log x = nt \log y \]

5. **Isolate \( t \)**:
Since \( t \) is common on both sides, we can cancel \( t \) (assuming \( t \neq 0 \)):
\[ m \log x = n \log y \]

6. **Differentiate with Respect to \( x \)**:
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(m \log x) = \frac{d}{dx}(n \log y) \]
Using the chain rule on the right side:
\[ \frac{m}{x} = \frac{n}{y} \frac{dy}{dx} \]

7. **Solve for \( \frac{dy}{dx} \)**:
Rearranging gives:
\[ \frac{dy}{dx} = \frac{m y}{n x} \]

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{m y}{n x} \]