1. **State the problem:** Solve the simultaneous equations: $$2\log y = \log 2 + \log x$$ $$2^{y} = 4^{x}$$ 2. **Rewrite the first equation:** Using logarithm property $\log a + \log b = \log (ab)$, $$2\log y = \log(2x)$$ 3. **Express it with a single log:** Since $2\log y = \log y^{2}$, $$\log y^{2} = \log(2x)$$ 4. **Equate inside arguments:** Because $\log a = \log b$ implies $a = b$, $$y^{2} = 2x$$ 5. **Rewrite the second equation:** Note that $4^{x} = (2^{2})^{x} = 2^{2x}$, so $$2^{y} = 2^{2x}$$ 6. **Equate exponents:** Since the bases are the same, $$y = 2x$$ 7. **Substitute $y=2x$ into $y^{2} = 2x$:** $$(2x)^{2} = 2x$$ $$4x^{2} = 2x$$ 8. **Simplify and solve for $x$:** $$4x^{2} - 2x = 0$$ $$2x(2x - 1) = 0$$ So, $$2x = 0 \implies x=0$$ $$2x - 1 = 0 \implies x=\frac{1}{2}$$ 9. **Find corresponding $y$ values:** Using $y=2x$, If $x=0$, then $y=0$. If $x=\frac{1}{2}$, then $y=1$. 10. **Check for validity:** Since $\log y$ appears, $y$ must be positive. For $y=0$ (not positive), discard this solution. For $y=1$ (positive), valid. **Final solution:** $$x=\frac{1}{2},\quad y=1$$