If`x^(y)=e^(x)," then "(logx)^(2)y_(1)=`

Options

A. `loge`
B. `logx`
C. `log(e//x)`
D. `log(x//e)`

Answer

To solve the problem where \( x^y = e^x \) and we need to find \( (\log x)^2 y_1 \), we will follow these steps:

### Step 1: Take the logarithm of both sides
We start with the equation:
\[ x^y = e^x \]
Taking the natural logarithm on both sides gives us:
\[ \log(x^y) = \log(e^x) \]

### Step 2: Apply logarithmic properties
Using the property of logarithms that states \( \log(a^b) = b \log(a) \), we can rewrite the equation:
\[ y \log x = x \log e \]
Since \( \log e = 1 \), this simplifies to:
\[ y \log x = x \]

### Step 3: Solve for \( y \)
From the equation \( y \log x = x \), we can isolate \( y \):
\[ y = \frac{x}{\log x} \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \) using the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \( u = x \) and \( v = \log x \). We find \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{1}{x} \).

Applying the quotient rule:
\[ y_1 = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \]

### Step 5: Find \( (\log x)^2 y_1 \)
Now we multiply \( y_1 \) by \( (\log x)^2 \):
\[ (\log x)^2 y_1 = (\log x)^2 \cdot \frac{\log x - 1}{(\log x)^2} \]
This simplifies to:
\[ (\log x)^2 y_1 = \log x - 1 \]

### Step 6: Express \( \log x - 1 \) in terms of logarithms
We can express \( \log x - 1 \) as:
\[ \log x - 1 = \log x - \log e = \log\left(\frac{x}{e}\right) \]

### Final Answer
Thus, we conclude that:
\[ (\log x)^2 y_1 = \log\left(\frac{x}{e}\right) \]