How to compute the left-inverse of $f$ in the sense of $\approx$? For two functions $\varphi,\psi:\Omega\times[0,\infty)\to[0,\infty)$, $\varphi\approx\psi$ means that there exist constants $c_1,c_2>0$, such that $\forall t\geq0$, $\forall x\in\Omega$, $$c_1\varphi(x,t)\leq\psi(x,t)\leq c_2\varphi(x,t).$$
for the function $f(x,t):=t^{p(x)}\ln(e+t)$, where $p(\cdot):\Omega\to(1,\infty)$ is a bounded function (this is an example appearing in papers about differential equations), then how to compute its $f^{-1}(x,t):=\inf\{u\geq0:f(x,u)\geq t\}$, $\forall x\in\Omega$, $t\geq0$ in the sense of $\approx$?
 A: The solution here mainly follows the lines of the previous answer. For brevity, write $p$ for $p(x)$. The condition that $p$ is bounded from above will not be used. We will only use the condition $p\ge1$.
If $f(t):=t^p\ln(e+t)$ for $p=p(x)\ge1$ and all real $t\ge0$, then
\begin{equation*}
    f^{-1}(u)\approx g(u):=\frac{u^{1/p}}{\ln^{1/p}(e+u^{1/p})} \tag{1}
\end{equation*}
for real $u\ge0$, where the symbol $\approx$ is used in the sense defined in your post.
Indeed, the function $f\colon[0,\infty)\to[0,\infty)$ is continuous and strictly increasing, from $f(0)=0$ to $f(\infty-):=\lim_{t\to\infty}f(t)=\infty$. So, $f^{-1}$ is the true inverse of $f$. That is, for all nonnegative real $t$ and $u$ we have
\begin{equation*}
    f(t)=u\iff f^{-1}(u)=t. \tag{2}
\end{equation*}
Next, for real $u\ge0$, let
\begin{equation*}
h(u):=u^{1/p}\,1(u\le e^2)+u^{1/(2p)}\,1(u>e^2) \tag{3}
\end{equation*}
and
\begin{equation*}
    r(u):=\frac{g(u)}{h(u)}.\tag{4} 
\end{equation*}
If $u\le e^2$, then
\begin{equation*}
    r(u)=\frac1{\ln^{1/p}(e+u^{1/p})}\ge\frac1{\ln^{1/p}(e+e^{2/p})}\ge\frac1{\ln(e+e^2)}>4/10, \tag{5} 
\end{equation*}
since $p\ge1$.
Consider now the case $u>e^2$. Let $d(u):=\ln(e+u^{1/p})-\sqrt{u}$. Then $d'(u)2p u (e+u^{1/p})=-e p \sqrt{u}+(2-p\sqrt{u})u^{1/p}\le-0+(2-\sqrt{e^2})u^{1/p}\le0$. So, $d$ is  decreasing on $[e^2,\infty)$ and $d(e^2):=\ln(e+e^{2/p})-e\le\ln(e+e^2)-e<0$. So, $d<0$ on $[e^2,\infty)$ and hence for $u>e^2$ we have $u^{1/(2p)}\ge\ln^{1/p}(e+u^{1/p})$ and hence
\begin{equation*}
    r(u)=\frac{u^{1/(2p)}}{\ln^{1/p}(e+u^{1/p})}\ge1. 
\end{equation*}
Thus, for all real $u\ge0$,
\begin{equation*}
    r(u)\ge K:=4/10. 
\end{equation*}
Therefore and by (1), for all real $u\ge0$,
\begin{equation*}
    Kh(u)\le g(u)\le u^{1/p}. \tag{6}
\end{equation*}
By the second inequality in (6),
\begin{equation*}
f(g(u))=u\,\frac{\ln(e+g(u))}{\ln(e+u^{1/p})}\le u=f(f^{-1}(u)). 
\end{equation*}
Therefore and because $f$ is strictly increasing, we have
\begin{equation*}
g(u)\le f^{-1}(u).\label{5} \tag{7}
\end{equation*}
On the other hand, by the first inequality in (6), for real $C\ge1$
\begin{equation*}
    f(Cg(u))=C^pu\,\frac{\ln(e+Cg(u))}{\ln(e+u^{1/p})} 
    \ge Cu R(u), \tag{8}
\end{equation*}
where
\begin{equation*}
R(u):=\frac{\ln(e+Kh(u))}{\ln(e+u^{1/p})}=\frac{\ln(e+K\tilde h(v))}{\ln(e+v^2)}, \tag{9}
\end{equation*}
where $v:=u^{1/(2p)}$ and $\tilde h(v):=h(v^{2p})\ge\min(v^2,v)$.
So, for all real $u\ge0$,
\begin{equation*}
R(u)\ge\rho(v):=\frac{\ln(e+K\min(v^2,v))}{\ln(e+v^2)}.
\end{equation*}
Note that $\rho$ is continuous and strictly positive on $[0,\infty)$, $\rho(0)=1>0$, and $\rho(v)\to1/2>0$ as $v\to\infty$. So, $\rho(v)\ge c$ for some real $c>0$ and all real $v\ge0$. So, $R(u)\ge c$ for all real $u\ge0$.
Taking now any real $C\ge1$ such that $Cc\ge1$, from (8) we get $f(Cg(u))\ge u=f(f^{-1}(u))$ for all real $u\ge0$, which implies $Cg(u)\ge f^{-1}(u)$.
Thus, in view of (7), $g(u)\le f^{-1}(u)\le Cg(u)$, for some universal real constant $C\ge1$. So, (1) is proved.
