1. **State the problem:** Solve for $x$ in the equation: $$\sqrt{\sqrt{(25^0)^3 + \frac{3}{4}} \div \frac{7}{256x}} = \sqrt{\sqrt{\frac{4x^2 + 256}{\left(\frac{120}{5!}\right)^{256}}}}$$ 2. **Simplify constants and expressions:** - Note that $25^0 = 1$ because any nonzero number to the zero power is 1. - Also, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. 3. **Rewrite the equation with these simplifications:** $$\sqrt{\sqrt{1^3 + \frac{3}{4}} \div \frac{7}{256x}} = \sqrt{\sqrt{\frac{4x^2 + 256}{\left(\frac{120}{120}\right)^{256}}}}$$ 4. **Simplify inside the roots:** - $1^3 = 1$ - $1 + \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{7}{4}$ - $\frac{120}{120} = 1$, so $1^{256} = 1$ So the equation becomes: $$\sqrt{\sqrt{\frac{7}{4}} \div \frac{7}{256x}} = \sqrt{\sqrt{4x^2 + 256}}$$ 5. **Simplify the left side division inside the nested root:** $$\sqrt{\frac{7}{4}} \div \frac{7}{256x} = \sqrt{\frac{7}{4}} \times \frac{256x}{7}$$ 6. **Simplify the multiplication:** $$= \frac{256x}{7} \times \sqrt{\frac{7}{4}} = \frac{256x}{7} \times \frac{\sqrt{7}}{2} = \frac{256x \sqrt{7}}{7 \times 2} = \frac{256x \sqrt{7}}{14}$$ 7. **Rewrite the entire equation:** $$\sqrt{\frac{256x \sqrt{7}}{14}} = \sqrt{\sqrt{4x^2 + 256}}$$ 8. **Square both sides to remove the outer square roots:** $$\frac{256x \sqrt{7}}{14} = \sqrt{4x^2 + 256}$$ 9. **Square both sides again to remove the remaining square root:** $$\left(\frac{256x \sqrt{7}}{14}\right)^2 = 4x^2 + 256$$ 10. **Calculate the left side:** $$\left(\frac{256x \sqrt{7}}{14}\right)^2 = \frac{(256)^2 x^2 \times 7}{14^2} = \frac{65536 x^2 \times 7}{196} = \frac{458752 x^2}{196}$$ 11. **Simplify the fraction:** $$\frac{458752}{196} = 2341.632653...$$ So: $$2341.632653 x^2 = 4x^2 + 256$$ 12. **Bring all terms to one side:** $$2341.632653 x^2 - 4x^2 = 256$$ $$2337.632653 x^2 = 256$$ 13. **Solve for $x^2$:** $$x^2 = \frac{256}{2337.632653} \approx 0.1095$$ 14. **Take the square root:** $$x = \pm \sqrt{0.1095} \approx \pm 0.331$$ **Final answer:** $$x \approx \pm 0.331$$