najdi x$$ \text{ }-\frac{1+\ln x}{x^2}=0 $$

We need to solve the equation:

\[
-\frac{1 + \ln x}{x^2} = 0
\]

To solve this equation, let's first understand what it means for the fraction on the left side to equal zero.

The fraction $-\frac{1 + \ln x}{x^2}$ will only be zero when the numerator $1 + \ln x$ is equal to zero, since the denominator $x^2$ is always positive for $x > 0$. Therefore, we set the numerator equal to zero:

\[
1 + \ln x = 0
\]

Now, solve for $x$:

\[
\ln x = -1
\]

Exponentiate both sides to remove the natural logarithm:

\[
x = e^{-1}
\]

So the solution is:

\[
x = \frac{1}{e}
\]

Would you like further details or have any questions?

Here are some related questions to explore further:

1. How do you solve equations involving natural logarithms?
2. What is the significance of the exponential function $e^x$ in solving logarithmic equations?
3. Can you find the derivative of the function $f(x) = 1 + \ln x$?
4. How does the domain of the natural logarithm affect the solution to this equation?
5. What are some other ways to solve the equation $1 + \ln x = 0$?

**Tip:** Remember, the natural logarithm $\ln x$ is only defined for $x > 0$.

Learn how to solve the equation -ln(x)/(x^2) = 0 step-by-step. Understand the concepts of logarithms, exponential functions, and equations. Suitable for high school students and anyone interested in algebraic equations involving logarithms and exponential functions.

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