A little game of $x$

$\ 2^x + 2^x + 2^x = \frac3{16}$
$or,~~\ 3 \times2^x = \frac3{16}$ $or,~~\ 2^x = \frac{1}{16}$
$or,~~\ 2^x = (\frac{1}{2})^4$
$or,~~\ 2^x = 2^{-4}$
$Since~~the~~ bases ~~are~~ same~~,~~ comparing ~~the~~powers$
$or,~~\boxed{\large x~=~-4}$

3( $2^x$ )= $\frac{3}{16}$

$2^x$ = $\frac{1}{16}$

$x=\log_2 \frac{1}{16}$

$x=-4$

In general for $2^x+2^x+2^x =\frac{3}{16}$ is:

$\begin{aligned}2^x+2^x+2^x =\frac{3}{16} &\Rightarrow 32^x= \frac{3}{16}\quad\quad\quad\quad\quad\quad\quad\quad{\text{Refine the } 2^x+2^x+2^x \space \text{into } 32^x} \\& \Rightarrow 482^x= 3 \\& \Rightarrow 2^x = \frac{1}{16} \quad\quad\quad\quad\quad\quad\quad\quad{\text{Apply the rule: If }a^{f(x)} = a^{g(x)} \space\text{, then } f(x)=g(x) } \\& \Rightarrow 2^x = 2^{-4} \\&\Rightarrow x = -4 \space\space\space\space \square \end{aligned}$

Other method for computation

\(\begin{aligned} 2^x+2^x+2^x =\frac{3}{16} &\Rightarrow 3(2^x) = \frac{3}{16} \\&\Rightarrow 2^x = \frac{3}{16 \cdot 3} \\&\Rightarrow 2^x=\frac{3}{48} \\&\Rightarrow 2^x = \frac{1}{16} \\&\Rightarrow 2^x = 2^{-4} \\&\Rightarrow x= -4 \space\space\space\space\space \square \end{aligned}

$\large \text{FIN!!!}$

3 X 2^x = 3/16.... 2^x = 1/16.... 16 X 2^x = 1.... 2^4 X 2^x = 1.... 2^{x+4} = 2^0.... x+4 = 0.... x = -4