Basic loggin'

This is quite a straightforward problem; just solve it conventionally:

${ 2 }^{ x }+{ 2 }^{ x+1 }+{ 2 }^{ x+2 }+{ 2 }^{ x+3 }={ 3 }^{ x }{ +3 }^{ x+1 }+{ 3 }^{ x+2 }+{ 3 }^{ x+3 }\\ 2^{ x }+2{ \times 2 }^{ x }+4{ \times 2 }^{ x }+8{ \times 2 }^{ x }={ 3 }^{ x }+{ 3\times 3 }^{ x }+9{ \times 3 }^{ x }+27{ \times 3 }^{ x }\\ 15\times { 2 }^{ x }=40\times { 3 }^{ x }\\ { \left( \frac { 2 }{ 3 } \right) }^{ x }=\frac { 8 }{ 3 } \\ \Longrightarrow x=\log _{ \frac { 2 }{ 3 } }{ \frac { 8 }{ 3 } }$

Now we've obtained our a and b: $a=\frac { 8 }{ 3 }$ , $b=\frac { 2 }{ 3 }$ .

$\rightarrow \sqrt { \frac { a }{ b } } =\sqrt { 4 } =2$

Therefore, $\boxed { \sqrt { \frac { a }{ b } } =2 }$ .

${2 }^{ x }+ { 2 }^{ x+1 }+{ 2 }^{ x+2 }+{ 2 }^{ x+3 }={ 3 }^{ x }{ +3 }^{ x+1 }+{ 3 }^{ x+2 }+{ 3 }^{ x+3 } \\ \Rightarrow 2^{x}(1+2+4+8)=3^{x}(1+3+9+27) \\ \Rightarrow \left(\frac{2}{3}\right)^{x}=\frac{8}{3} \Rightarrow x=\log_{\frac{2}{3}}\frac{8}{3}=\log_{b}{a} \\ \Rightarrow \sqrt{\frac{a}{b}}=\sqrt{\frac{\frac{8}{3}}{\frac{2}{3}}}=\sqrt{4}=\boxed{2}$

Use multiplication, division and change of base properties of logarithm