Logarithmic Warmup

$\log _{ 4 }{ \quad x } =a\\ \quad \quad \quad x=4^{ a }={ 2 }^{ 2a }$

$\log _{ 2 }{ \quad y } =b\\ \quad \quad \quad y={ 2 }^{ b }$

Multiply $x$ by $y$ :

$xy=2^{ 2a }\times 2^{ b }=2^{ 2a+b }$

Divide $x$ by $y$ :

$\frac { x }{ y } =\frac { 2^{ 2a } }{ 2^{ b } } =2^{ 2a-b }$

Since, $xy=128$ and $\frac { x }{ y }=4$ ,

$2^{ 2a+b }=128=2^{ 7 }\\ 2^{ 2a-b }=\quad 4 =2^{ 2 }\\$

Thus,

$2a+b=7\longrightarrow (1)\\2a-b=2\longrightarrow (2)$

Adding $(1)$ to $(2)$ ,

$\quad 4a=9\\ \Rightarrow a=2.25\\$

Substituting $a=2.25$ into $(1)$ , we get $b= 2.5$ .

Since $x=4^{ a }$ and $y=2^{ b }$ ,

$a+b+x+y=2.25+2.5+4^{ 2.25 }+2^{ 2.5 }\approx \boxed { 33.03 } \\$ _(Correct to 2 dec. places) _

$\begin{cases} \log_4 x = a & \implies x = 4^a = 2^{2a} \\ \log_2 y = b & \implies y = 2^b \end{cases}$

$\implies \begin{cases} xy = 2^{2a+b} = 128 = 2^7 & \implies 2a+b = 7 & ...(1) \\ \dfrac xy = 2^{2a-b} = 4 = 2^2 & \implies 2a-b = 2 & ...(2) \end{cases}$

$\begin{aligned} (1)+(2): \quad 4a & = 9 & \implies a = \frac 94 \\ (1): \quad \frac 92 + b & = 7 & \implies b = \frac 52 \\ \implies x & = 2^\frac 92 \\ y & = 2^\frac 52 \end{aligned}$

$\implies a+b+x+y \approx \boxed{33.03}$