Have fun L'hopitaling this one!

As $x \to 0$ , $1 - \cos x \approx \dfrac{1}{2} x^2$ and $\ln(1+x) \approx x$ . This comes from each respective Taylor polynomial. We can apply these two formulas in this limit.

$= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{\ln^{116}\left(2x + 1 + 2 - 2 \cos x \right) }$ $= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{\ln^{116}\left(2x + 1 + 2( 1- \cos x) \right) }$ $= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{\ln^{116}(2x + 1 + x^2 ) }$ $= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{\ln^{116}( x+1)^2 }$ $= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{2 \ln^{116}( x+1) }$
$= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{2 x^{116} }$
$= \displaystyle \lim_{x \to 0} \dfrac{\dfrac{1}{2^{58}}x^{116}}{2 x^{116} }$
$L = \dfrac{1}{2^{59}}; \log_{2} L = \boxed{-59}$

$\begin{aligned} \lim_{x \to 0} \dfrac{(1-\cos x)^{58}}{\ln^{116}(2x+3-2\cos x)} &= \left(\lim_{x \to 0} \dfrac{1-\cos x}{\ln^2(2x+3-2\cos x)}\right)^{58} \\ &= \left(\lim_{x \to 0} \dfrac{2\sin^2 \left(\frac{x}{2}\right)}{\ln^2(2x+3-2\cos x)}\right)^{58} \\ &= \left(2\lim_{x \to 0} \left( \dfrac{\sin \left(\frac{x}{2}\right)}{\ln(2x+3-2\cos x)} \right)^2 \right)^{58} \\ &= \left(2\lim_{x \to 0} \left( \dfrac{\dfrac{1}{2} \cos \left(\frac{x}{2}\right)}{\frac{2+2\sin x}{2x+3-2\cos x}} \right)^2 \right)^{58} \\ &= \left(2\left(\dfrac{1}{4}\right)^2\right)^{58}\\ &= \left(\dfrac{1}{8}\right)^{58}\\ &= 2^{\boxed{-174}} \end{aligned}$

$\begin{aligned} L & = \lim_{x \to 0} \frac {(1-\cos x)^{58}}{\ln^{116}(2x+3-2\cos x)} \\ & = \lim_{x \to 0} \left(\frac {1-\cos x}{\ln^2(2x+3-2\cos x)}\right)^{58} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \left(\frac {\sin x}{ \frac {2 \ln(2x+3-2\cos x)(2+\sin x)}{2x+3-2\cos x}}\right)^{58} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to 0} \left(\frac {\sin x(2x+3-2\cos x)}{2 \ln(2x+3-2\cos x)(2+\sin x)}\right)^{58} & \small \color{#3D99F6} \text{A 0/0 case again.} \\ & = \lim_{x \to 0} \left(\frac {\cos x(2x+3-2\cos x)+\sin x(2+2\sin x)}{\frac {2 (2+\sin x)^2}{2x+3-2\cos x} + 2 \ln(2x+3-2\cos x)\cos x}\right)^{58} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \left(\frac 18\right)^{58} = 2^{-174} \end{aligned}$

$\implies \log_2 L = \boxed{-174}$ .