1. **State the problem:** Solve the equation $$\log_{10} \frac{x^2 - 24}{x} = 1$$. 2. **Recall the definition of logarithm:** If $$\log_b A = C$$, then $$A = b^C$$. 3. **Apply the definition:** Here, base $$b = 10$$ and $$C = 1$$, so $$\frac{x^2 - 24}{x} = 10^1 = 10$$. 4. **Rewrite the equation:** $$\frac{x^2 - 24}{x} = 10$$ 5. **Multiply both sides by $$x$$ to clear the denominator:** $$\cancel{x} \cdot \frac{x^2 - 24}{\cancel{x}} = 10x$$ which simplifies to $$x^2 - 24 = 10x$$. 6. **Bring all terms to one side to form a quadratic equation:** $$x^2 - 10x - 24 = 0$$. 7. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-10$$, and $$c=-24$$. 8. **Calculate the discriminant:** $$\Delta = (-10)^2 - 4 \times 1 \times (-24) = 100 + 96 = 196$$. 9. **Find the roots:** $$x = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2}$$. 10. **Calculate each root:** - $$x = \frac{10 + 14}{2} = \frac{24}{2} = 12$$ - $$x = \frac{10 - 14}{2} = \frac{-4}{2} = -2$$ 11. **Check the domain restrictions:** - The argument of the logarithm must be positive: $$\frac{x^2 - 24}{x} > 0$$. - Also, $$x \neq 0$$. 12. **Test $$x=12$$:** $$\frac{12^2 - 24}{12} = \frac{144 - 24}{12} = \frac{120}{12} = 10 > 0$$, valid. 13. **Test $$x=-2$$:** $$\frac{(-2)^2 - 24}{-2} = \frac{4 - 24}{-2} = \frac{-20}{-2} = 10 > 0$$, but since $$x=-2$$ is negative and the denominator is negative, the fraction is positive, so it is valid. 14. **Final answer:** Both $$x=12$$ and $$x=-2$$ satisfy the equation and domain restrictions. **Answer:** $$x = 12 \text{ or } x = -2$$.