1. **State the problem:** Solve the equation $\left(\log_3 x\right)^2 - 3\log_3 x = 10$ for $x$. 2. **Use substitution:** Let $y = \log_3 x$. The equation becomes: $$y^2 - 3y = 10$$ 3. **Rewrite the equation:** Move all terms to one side: $$y^2 - 3y - 10 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-3$, and $c=-10$. Calculate the discriminant: $$\Delta = (-3)^2 - 4 \times 1 \times (-10) = 9 + 40 = 49$$ Calculate the roots: $$y = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}$$ So, - $y_1 = \frac{3 + 7}{2} = 5$ - $y_2 = \frac{3 - 7}{2} = -2$ 5. **Back-substitute to find $x$:** Recall $y = \log_3 x$, so For $y_1 = 5$: $$\log_3 x = 5 \implies x = 3^5 = 243$$ For $y_2 = -2$: $$\log_3 x = -2 \implies x = 3^{-2} = \frac{1}{9}$$ 6. **Check domain:** Since $x > 0$ for logarithms, both $x=243$ and $x=\frac{1}{9}$ are valid. **Final answer:** $$x = 243 \text{ or } x = \frac{1}{9}$$