Don't You Wish They Were Ones

$\Large\color{#D61F06}{\log_2}( 2 \color{#D61F06}{\times} 2 \color{#D61F06}{\times} 2 \color{#D61F06}{\times} 2 \color{#D61F06}{\times}2 ) = \ 5$ $\textbf{OR}$ $\Large ((2 \color{#20A900}{\times} 2 \color{#20A900}{\times} 2 )\color{#20A900}{+} 2 )\color{#20A900}{÷}2 ) = \ 5$ $\textbf{OR}$ $\Large 2 \color{#0C6AC7}{+} 2 \color{#0C6AC7}{+} 2 \color{#0C6AC7}{-} (2 \color{#0C6AC7}{-}2 )\color{#0C6AC7}{!} = \ 5$ $\textbf{OR}$ $\Large \color{#EC7300}{\lfloor \left(\sqrt{\color{#333333}{2 \color{#EC7300}{\times} 2 \color{#EC7300}{\times} 2 \color{#EC7300}{\times} 2 \color{#EC7300}{\times}2 }}\right)\rfloor} = \ 5$ $\textbf{OR}$ $\huge \color{#007fff}{\lceil \color{#333333}{2\color{#007fff}{.}222}\rceil}\color{#007fff}{+}2=5$ $\textbf{OR}$ $\Large 2\color{#ff00cc}{+}2\color{#ff00cc}{+}\color{#ff00cc}{(}\color{#ff00cc}{(}2\color{#ff00cc}{-}2\color{#ff00cc}{)}\color{#ff00cc}{\times}2\color{#ff00cc}{)!}=5$ $\textbf{OR}$ $\Large \color{forestgreen}{\int_0^5 (}2\color{forestgreen}{\times}2\color{forestgreen}{\times}\color{forestgreen}{(}2\color{forestgreen}{-}2\color{forestgreen}{)}\color{forestgreen}{\times}2\color{forestgreen}{)!~dx}=5$

2+2+2-2/2=5

$\begin{aligned} \Huge{\color{#D61F06}{2+2+2-\dfrac{2}{2}=5}}\\ \Huge{\color{#3D99F6}{(2+2)!\div (2+2)-\phi(2)=5}}\\ \Huge{\color{#20A900}{\phi(2)+\dfrac{2}{2\div(2+2)}=5}}\\ \Huge{\color{#EC7300}{2+2+\left\lceil\dfrac{2}{2+2}\right\rceil=5}}\\ \Huge{\color{#624F41}{2^2+\phi(2)\left(\dfrac{2}{2}\right)=5}}\\ \Huge{\color{#CEBB00}{\left\lfloor\dfrac{(2+2)!-2}{2+2}\right\rfloor=5}}\\ \Huge{\color{#E81990}{\left\lceil\sqrt{2^{2^2}+\dfrac{2}{2}}\right\rceil=5}} \end{aligned}$

$\Huge \color{#3D99F6}{\phi(2)+\phi(2)+\phi(2)+\phi(2)+\phi(2)=5}$

Note: Hint is there is the title :)

$(2+2/2)!-(2/2)$

2-(2/2)+(2*2) = 5

$2+2+2-2:2=5$

$2^{2-2}+2^2$

(2:2)+(2x2):2=5

2x2 + 2^(2 - 2) = 5 OR 2^2 + 2^(2 - 2) = 5

((2×2)+2)-(2÷2)=5

I got lazy and copied my primary school textbook. (2x2x2+2)/2 :D

(2 raised to 0)X2

Yes this is possible: ((2×2×2)+2)÷2)

And partially because NOTHING'S IMPOSSIBLE :P

{[(2×2)! / 2 ] - 2 } / 2 = 5

$\log _{ { 2 }^{ { 2 }^{ -2 } } }{ 2 } +\phi (2)$

(2+2+2) - (2/2) = 5

log(2^2^2 * 2)/log(2)

( ((( 2 / 2 ) / 2 ) + 2 ) * 2 )

$[\sqrt{2}]+[\sqrt{2}]+[\sqrt{2}]+[\sqrt{2}]+[\sqrt{2}]$ , Did I do it right...?

(2+2÷2)! - 2÷2

This could be the operation : 2+2+2-2/2 = 2+2+2-1= 6-1 = 5

Keep it simple

2 × 2 + 2 - 2 ÷ 2 = 4 + 2 - 1 = 5

(22-2)/(2+2)=5

{2^(2-2)} +2+2=5

2^0 + 2^0 + 2^0 + 2^0 + 2^0 = 5

2^(2-2) +2+2

2 × 2 - 2 ÷ 2 + 2 = 5

2-(2/2)+2+2

(2/2/2+2)*2

Could the operation be the number of 2 or the number of operands?

2+2+2-(2÷2)=5

2^2+2-2÷2=5

$\sqrt{2×2}+2+(2/2) = 5$