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<sRGB> specifies a red-green-blue colour encoding. Restricting attention to component values in $[0,1]$, its encoded component $s$ corresponds to the linear-light component
$$
D(s)=\begin{cases}s/12.92,&0\leq s\leq0.04045,\\((s+0.055)/1.055)^{2.4},&0.04045<s\leq1.\end{cases}
$$
Thus the conversion is piecewise, not a single power law. The reverse conversion used here is
$$
E(l)=\begin{cases}12.92l,&0\leq l\leq0.0031308,\\1.055l^{1/2.4}-0.055,&0.0031308<l\leq1.\end{cases}
$$
These conventional rounded thresholds are used by the W3C conversion routines; they give a tiny mismatch at the join rather than a mathematically exact pair of inverses everywhere.https://www.w3.org/TR/css-color-4/#predefined-sRGB[W3C, CSS Color Module Level 4, section 10.2: encoded and linear-light sRGB]{ref}https://www.w3.org/TR/css-color-4/#color-conversion-code[W3C, CSS Color Module Level 4, sample colour-conversion code: lin_sRGB and gam_sRGB]{ref}

For an eight-<bit> neutral <grayscale> code $q$, use $s=q/255$ in all three components. Its <relative luminance> is then $D(s)$. For example, $D(0.5)\approx0.21404$, whereas $E(0.5)\approx0.73536$. This is why averaging encoded values and averaging linear light give different answers. <sRGB> is a specified interpretation; merely calling a <raster image> <grayscale> does not imply it uses <sRGB>.


 sRGB (source code)