Solve using logarithm and explation

The first equation from the image is $2\log_y x + 2\log_x y = 5$.
Let $u = \log_y x$.
Using the logarithm property $\log_a b = \frac{1}{\log_b a}$, we can write $\log_x y = \frac{1}{\log_y x} = \frac{1}{u}$.
Substitute these into the equation:

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

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

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

Rearrange the equation into the standard quadratic form $au^2 + bu + c = 0$:

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

Factor the quadratic equation:

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

This gives 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 $\log_y x = \frac{1}{2}$ or $\log_y x = 2$. This completes the first part of the question.

We consider two cases based on the values of $\log_y x$, along with the second given equation $xy = 27$.

Case 1: $\log_y x = 2$
From the definition of logarithm, this means $x = y^2$.
Substitute $x = y^2$ into the second 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 $(x, y) = (9, 3)$.

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

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

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:

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