1. **Solve** $\frac{x^2}{3} = 9$. Multiply both sides by 3: $$x^2 = 27$$ Take the square root: $$x = \pm \sqrt{27} = \pm 3\sqrt{3}$$ 2. **Solve** $x = 9 \sqrt{9x^{1/2}}$. Rewrite the square root: $$x = 9 \sqrt{9} \sqrt{x^{1/2}} = 9 \times 3 \times x^{1/4} = 27 x^{1/4}$$ Divide both sides by $x^{1/4}$ (assuming $x>0$): $$x^{3/4} = 27$$ Raise both sides to the power $\frac{4}{3}$: $$x = 27^{4/3} = (3^3)^{4/3} = 3^4 = 81$$ 3. **Solve** $8^{2x + 1} = 243$. Express bases as powers of primes: $$8 = 2^3, \quad 243 = 3^5$$ Rewrite: $$(2^3)^{2x+1} = 3^5 \implies 2^{6x+3} = 3^5$$ Since bases differ, no real solution unless complex logs are used. Using logarithms: $$ (6x+3) \ln 2 = 5 \ln 3 \implies 6x+3 = \frac{5 \ln 3}{\ln 2}$$ Solve for $x$: $$x = \frac{1}{6} \left( \frac{5 \ln 3}{\ln 2} - 3 \right)$$ 4. **Solve** $5^x = \frac{125}{\sqrt{5}}$. Rewrite right side: $$125 = 5^3, \quad \sqrt{5} = 5^{1/2}$$ So: $$5^x = \frac{5^3}{5^{1/2}} = 5^{3 - 1/2} = 5^{5/2}$$ Equate exponents: $$x = \frac{5}{2}$$