$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Consider $f(x)=x\ln(2) - 61\ln(x+2) -200$ for $x>-2$.
It has one minimum at $x=-2+\tfrac{61}{\ln2}$,
so in order to solve $f(x)>0$,
you just need to find two zeros of $f(x)$
in terms of the Lambert W function, which can be done as follows:
\begin{align}
x\ln(2) - 61\ln(x+2) -200&=0
,\\
(x+2)\ln2-2\ln2 - 61\ln(x+2)-200&=0
,\\
(x+2)\ln2-61\ln(x+2)&=200+2\ln2
,\\
-\tfrac{\ln2}{61}(x+2)+\ln(x+2)
&= -\tfrac{200}{61}-\tfrac2{61}\ln2
,\\
\ln(\exp(-\tfrac{\ln2}{61}(x+2)))+\ln(x+2)
&=-\tfrac{200}{61}-\tfrac2{61}\ln2
,\\
\ln((x+2)\exp(-\tfrac{\ln2}{61}(x+2)))
&=-\tfrac{200}{61}-\tfrac2{61}\ln2
,\\
(x+2)\exp(-\tfrac{\ln2}{61}(x+2))
&=\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)
,\\
-\tfrac{\ln2}{61}(x+2)\exp(-\tfrac{\ln2}{61}(x+2))
&=-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)
,\\
\W\left(-\tfrac{\ln2}{61}(x+2)\exp(-\tfrac{\ln2}{61}(x+2)) \right)
&=\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right)
,\\
-\tfrac{\ln2}{61}(x+2)
&=\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right)
,\\
x
&=-2-\tfrac{61}{\ln2}\W\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right)
.
\end{align}
Now, as we have the solution of $f(x)=0$ in terms of the Lambert W function,
we need to check its argument
\begin{align}
z&=-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)
\approx -0.0004185.
\end{align}
As we can see, $z\in (-\tfrac1\e,0)$, so indeed we have two solutions
\begin{align}
x_0
&=
-2-\tfrac{61}{\ln2}\Wp\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right)
\approx -1.963153548947
,\\
\text{and }\quad
x_1
&=-2-\tfrac{61}{\ln2}\Wm\left(-\tfrac{\ln2}{61}\exp(-\tfrac{200}{61}-\tfrac2{61}\ln2)\right)
\approx 885.9981973778
.
\end{align}
Thus, the answer is
\begin{align}
f(x)&>0
\quad \text{for }\quad x\in (-2,x_0) \cup x\in(x_1,\infty)
.
\end{align}
$\endgroup$