Solve the following simultaneous equations. a) ylog, 8 = x and 2x+8y = 8192 (b) log, x+2log, y=4 and x+12y=52.

The first equation is $2\log_y x + 2\log_x y = 5$.
Let $u = \log_y x$.
Using the change of base property of logarithms, $\log_x y = \frac{1}{\log_y x}$, so $\log_x y = \frac{1}{u}$.
Substitute these into the first equation:

$$2u + \frac{2}{u} = 5$$

Multiply the entire equation by $u$ (assuming $u \neq 0$):

$$2u^2 + 2 = 5u$$

Rearrange into standard quadratic form:

$$2u^2 - 5u + 2 = 0$$

Factor the quadratic equation:

$$(2u - 1)(u - 2) = 0$$

This yields two possible values for $u$:

$$2u - 1 = 0 \implies u = \frac{1}{2}$$ $$u - 2 = 0 \implies u = 2$$

Since $u = \log_y x$, we have shown that $\log_y x = \frac{1}{2}$ or $\log_y x = 2$.

$\log_y x = 2$.
From the definition of logarithm, $\log_y x = 2$ means $x = y^2$.
Substitute this into the second given equation, $xy = 27$:

$$(y^2)y = 27$$ $$y^3 = 27$$

Take the cube root of both sides:

$$y = \sqrt[3]{27}$$ $$y = 3$$

Now substitute $y = 3$ back into $x = y^2$:

$$x = 3^2$$ $$x = 9$$

This gives the pair $(x, y) = (9, 3)$.

$\log_y x = \frac{1}{2}$.
From the definition of logarithm, $\log_y x = \frac{1}{2}$ means $x = y^{1/2} = \sqrt{y}$.
Substitute this into the second given equation, $xy = 27$:

$$(\sqrt{y})y = 27$$ $$y^{1/2} \cdot y^1 = 27$$ $$y^{3/2} = 27$$

To solve for $y$, raise both sides to the power of $\frac{2}{3}$:

$$y = (27)^{2/3}$$ $$y = (\sqrt[3]{27})^2$$ $$y = 3^2$$ $$y = 9$$

Now substitute $y = 9$ back into $x = \sqrt{y}$:

$$x = \sqrt{9}$$ $$x = 3$$

This gives the pair $(x, y) = (3, 9)$.

The pairs of values for $x$ and $y$ that satisfy both equations are:

$$\boxed{(9, 3) \text{ and } (3, 9)}$$