1. **State the problem:** Solve the equation $$x^2 + \log x + 24 = 0$$ for $x$. 2. **Important note:** The logarithm function $\log x$ is defined only for $x > 0$. 3. **Rewrite the equation:** We want to find $x > 0$ such that $$x^2 + \log x + 24 = 0.$$ 4. **Analyze the equation:** Since $x^2$ and 24 are positive for all $x \neq 0$, and $\log x$ grows slowly, let's check if there are any real positive solutions. 5. **Check behavior:** For $x$ close to 0, $\log x \to -\infty$, so the left side tends to $+\infty$ (because $x^2$ is near 0 and 24 is positive, but $\log x$ is very negative). For large $x$, $x^2$ dominates and the left side is positive. 6. **Try to find roots numerically:** - At $x=1$, $1 + 0 + 24 = 25 > 0$ - At $x=0.1$, $0.01 + \log 0.1 + 24 = 0.01 - 1 + 24 = 23.01 > 0$ 7. Since the expression is always positive for $x > 0$, there are no real solutions. **Final answer:** There is **no real solution** to the equation $x^2 + \log x + 24 = 0$ for $x > 0$.