Simplify it all

Let's convert the logarithms to base 2:

$\frac { 2\log _{ 2 }{ \left( 2y \right) } }{ \log _{ 2 }{ \left( x \right) } } =\frac { 2\log _{ 2 }{ \left( 4z \right) } }{ \log _{ 2 }{ \left( 2x \right) } } =\frac { \log _{ 2 }{ \left( 8yz \right) } }{ \log _{ 2 }{ \left( 2{ x }^{ 4 } \right) } } \neq 0\\ \frac { 2\log _{ 2 }{ \left( 2 \right) } +2\log _{ 2 }{ \left( y \right) } }{ \log _{ 2 }{ \left( x \right) } } =\frac { 2\log _{ 2 }{ \left( 4 \right) } +2\log _{ 2 }{ \left( 4 \right) } }{ \log _{ 2 }{ \left( 2 \right) } +\log _{ 2 }{ \left( x \right) } } =\frac { \log _{ 2 }{ \left( 8 \right) } +\log _{ 2 }{ \left( y \right) } +\log _{ 2 }{ \left( z \right) } }{ \log _{ 2 }{ \left( 2 \right) } +4\log _{ 2 }{ \left( x \right) } } \neq 0$

Let $a=\log _{ 2 }{ \left( x \right) } ,b=\log _{ 2 }{ \left( y \right) } ,c=\log _{ 2 }{ \left( z \right) }$ . Then:

$\frac { 2+2b }{ a } =\frac { 4+2c }{ 1+a } =\frac { 3+b+c }{ 1+4a } \neq 0$ , meaning $b\neq -1$

Now we've a system with two equations:

$\begin{cases} \left( 1+a \right) \cdot 2\left( 1+b \right) =2\left( 2+c \right) \cdot a \\ \left( 1+4a \right) \cdot 2\left( 1+b \right) =\left( 3+b+c \right) \cdot a \end{cases}\\ \begin{cases} 1+a+b+ab=2a+ac \\ 2+8a+2b+8ab=3a+ab+ac \end{cases}\\ \begin{cases} ac=1-a+b+ab \\ ac=2+5a+2b+7ab \end{cases}\\ 2+5a+2b+7ab=1-a+b+ab\\ 1+6a+b+6ab=0\\ \left( 1+6a \right) \cdot \left( 1+b \right) =0$

Since $b\neq -1$ , we know $a=-\frac { 1 }{ 6 }$ :

$-\frac { 1 }{ 6 } \cdot c=1+\frac { 1 }{ 6 } +b-\frac { 1 }{ 6 } \cdot b\\ c=-7-5b$

Now we can get the answer:

$x\cdot { y }^{ 5 }\cdot z={ 2 }^{ a+5b+c }={ 2 }^{ -\frac { 1 }{ 6 } +5b-7-5b }={ 2 }^{ -\frac { 1 }{ 6 } -7 }={ 2 }^{ -\frac { 43 }{ 6 } }=\frac { 1 }{ { 2 }^{ \frac { 43 }{ 6 } } }$

$p=43,q=6\Rightarrow p+q=\boxed { 49 }$

Step 1: Let $n = 2 \log_{x}{ (2y) } = 2 \log_{2x}{ (4z) } = \log_{2x^4}{ (8yz) } \not = 0$

Step 2: Convert each of the expressions to an exponential for our simplification

Expression 1: $n = 2 \log_{x}{ (2y) } \implies x^{n/2} = 2y$

Expression 2: $n = 2 \log_{2x}{ (4z) } \implies (2x)^{n/2} = 4z$

Expression 3: $n = \log_{2x^4}{ (8yz) } \implies (2x^4)^{n} = 8yz$ $\color{#3D99F6} \text{ [Note: E1 } \times \text{ E2 = E3 ]}$

Step 3: Thus as E1 $\times$ E2 = E3

$\implies$ Products of E1 $\times$ E2 = E3

$\implies 2^{n/2} \cdot x^n = 2^n x^{4n} \implies \color{#EC7300} x^{3n} = 2^{-n/2}$ $\implies x = 2^{-1/6}$

Step 4: Lets get back to the question :)

We have substituted the "expressions" and raised it to the required power.

$xy^5 z = 2^{-1/6} \times (x^{5n/2} \cdot 2^{-5} ) \times (2^{n/2} \cdot x^{n/2} \cdot 2^{-2} )$ $\small \color{#20A900} \text{ For easiness of simplification I have separated the bases of powers }$

$xy^5 z = 2^{-43/6} \times x^{3n} \times 2^{n/2} \small \color{#EC7300} \text{ Notice a similarity with the above orange expression} \therefore \text {this is = 1 }$

Step 5: Find the value of $p + q$

Hence: $xy^5 z = 2^{-43/6} = \dfrac{1}{2^{43/6}}$ and by substitution $p + q = \boxed{ 49 }$

Let $2 \log_x (2y) = 2 \log_{2x} (4x) = \log_{2x^4} (8yz) = 2k$ . Using a log base of 2, we have:

$\begin{cases} \dfrac {\log_2(2y)}{\log_2 x} = k & \implies \log_2 (2y) = k \log_2 x & ...(1) \\ \dfrac {\log_2(4z)}{\log_2 (2x)} = k & \implies \log_2 (4z) = k \log_2 (2x) & ...(2) \\ \dfrac {\log_2(8yz)}{\log_2 (2x^4)} = 2k & \implies \log_2 (2y) + \log_2 (4z) = 2k \log_2 (2x^4) & ...(3) \end{cases}$

From $(1)+(2) = (3)$ :

$\begin{aligned} k \log_2 x + k \log_2(2x) & = 2k \log_2(2x^4) & \small \color{#3D99F6} \text{Since }k \ne 0 \\ \log_2 x + \log_2(2x) & = 2 \log_2(2x^4) \\ \log_2(2x^2) & = \log_2(4x^8) \\ 2x^2 & = 4x^8 \\ x^6 & = \frac 12 \\ \implies x & = \frac 1{2^\frac 16} \end{aligned}$

From $(1) \div (2)$ :

$\begin{aligned} \frac {\log_2 (2y)}{\log_2 (4z)} & = \frac {\log_2 x}{1 + \log_2 x} = \frac {-\frac 16}{1-\frac 16} = - \frac 15 & \small \color{#3D99F6} \text{Since }\log_2 x = - \frac 16 \\ 5 \log_2 (2y) & = - \log_2 (4z) \\ 5 \log_2 (2y) + \log_2 (4z) & = 0 \\ \log_2 (2^5y^5 \cdot 2^2 z) & = 0 \\ 2^7y^5z & = 1 \\ y^5z & = \frac 1{2^7} \\ \implies xy^5z & = \frac 1{2^{7+\frac 16}} = \frac 1{2^\frac {43}6} \end{aligned}$

Therefore, $p+q = 43+6 = \boxed{49}$ .