1. **State the problem:** Solve the simultaneous equations: $$2\log y = \log 2 + \log x$$ and $$2^y = 4^x$$ 2. **Rewrite the first equation using log properties:** Recall that $\log a + \log b = \log(ab)$, so $$2\log y = \log(2) + \log(x) = \log(2x)$$ Also, $2\log y = \log y^2$, so $$\log y^2 = \log 2x$$ 3. **Equate the arguments of the logarithms:** Since $\log y^2 = \log 2x$, we have $$y^2 = 2x$$ 4. **Rewrite the second equation:** Note that $4^x = (2^2)^x = 2^{2x}$, so $$2^y = 2^{2x}$$ 5. **Equate the exponents since bases are equal:** $$y = 2x$$ 6. **Substitute $y = 2x$ into $y^2 = 2x$:** $$ (2x)^2 = 2x $$ $$ 4x^2 = 2x $$ 7. **Solve for $x$:** Divide both sides by $x$ (assuming $x \neq 0$): $$4x = 2$$ $$x = \frac{2}{4} = \frac{1}{2}$$ 8. **Find $y$ using $y = 2x$:** $$y = 2 \times \frac{1}{2} = 1$$ 9. **Check the solution:** - Check first equation: $$2\log y = 2\log 1 = 2 \times 0 = 0$$ $$\log 2 + \log x = \log 2 + \log \frac{1}{2} = \log(2 \times \frac{1}{2}) = \log 1 = 0$$ - Check second equation: $$2^y = 2^1 = 2$$ $$4^x = 4^{\frac{1}{2}} = \sqrt{4} = 2$$ Both equations hold true. **Final answer:** $$x = \frac{1}{2}, \quad y = 1$$