1. **State the problem:** We need to prove that the equation $g(x) = x^2 + 2\ln x = 0$ has exactly one solution for $x > 0$. 2. **Recall the domain and function:** The function $g(x) = x^2 + 2\ln x$ is defined for $x > 0$ because $\ln x$ is only defined for positive $x$. 3. **Analyze the behavior of $g(x)$:** - As $x \to 0^+$, $\ln x \to -\infty$, so $g(x) \to -\infty$. - As $x \to \infty$, $x^2$ dominates and $g(x) \to \infty$. 4. **Find the derivative to check monotonicity:** $$g'(x) = 2x + \frac{2}{x} = 2\left(x + \frac{1}{x}\right)$$ 5. **Analyze the derivative:** - Since $x > 0$, both $x$ and $\frac{1}{x}$ are positive. - Therefore, $g'(x) > 0$ for all $x > 0$. 6. **Interpretation:** - $g(x)$ is strictly increasing on $(0, \infty)$. 7. **Check values to confirm root existence:** - At $x=1$, $g(1) = 1^2 + 2\ln 1 = 1 + 0 = 1 > 0$. - At $x=0.1$, $g(0.1) = (0.1)^2 + 2\ln(0.1) = 0.01 + 2(-2.302585) = 0.01 - 4.60517 = -4.59517 < 0$. 8. **Conclusion:** - Since $g(x)$ is continuous and strictly increasing from $-\infty$ to $\infty$ on $(0, \infty)$ and changes sign from negative to positive, it crosses zero exactly once. **Final answer:** The equation $g(x) = 0$ has exactly one solution for $x > 0$.