Solve the following system of equations: 2 x + 24 y = 4 and 10 (2x+y) = 1. Show that xy = 16.

$$\log_4 y = \frac{\log_2 y}{\log_2 4}$$

Since $2^2 = 4$, we have $\log_2 4 = 2$.

$$\log_4 y = \frac{\log_2 y}{2}$$

$$\log_2 x + 2\left(\frac{\log_2 y}{2}\right) = 4$$ $$\log_2 x + \log_2 y = 4$$

$$xy = 2^4$$ $$xy = 16$$

This completes the first part of the problem.

From Part 1, we know that $\log_2 x + 2\log_4 y = 4$ simplifies to:

$$xy = 16 \quad (*)$$

$$\log_{10} (2x+y) = 1$$

Convert this to an exponential equation:

$$2x+y = 10^1$$ $$2x+y = 10 \quad (**)$$

From equation $(**)$, express $y$ in terms of $x$:

$$y = 10 - 2x$$

$$x(10 - 2x) = 16$$ $$10x - 2x^2 = 16$$

$$2x^2 - 10x + 16 = 0$$

Divide the entire equation by 2:

$$x^2 - 5x + 8 = 0$$

Here, $a=1$, $b=-5$, $c=8$.
Calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (-5)^2 - 4(1)(8)$$ $$\Delta = 25 - 32$$ $$\Delta = -7$$

Since the discriminant $\Delta = -7$ is negative, the quadratic equation has no real solutions for $x$.
For logarithms to be defined in the real number system, their arguments ($x$ and $y$) must be positive real numbers. Since there are no real values for $x$, there are no real values for $y$ either.

Therefore, there are no real solutions for $x$ and $y$ that satisfy the given system of equations.

The problem has no real solutions for $x$ and $y$.