With substitution $z=e^x$ you get the following equation
$$\sqrt{2-3 z^2}-z=2 z^c$$
which I think is generally unsolvable in closed analytical form $z(c)$ and hence $x(c)$ for arbitrary $c$. Neither Solve
(which "deals primarily with linear and polynomial equations." see details section in docs)
nor Reduce
can help due to principal math limitations.
Then you are left with two choices:
- Exact solutions for some integer and rational $c$
- Numeric solution
Some integer and rational $c$
$c=1/3$
Solve[Log[1/2 (-E^x + Sqrt[2 - 3 E^(2 x)])] == x/3, x]
$x=-\frac{1}{2} \log \left(\frac{6}{7-\frac{10}{\sqrt[3]{53-3 \sqrt{201}}}-\sqrt[3]{53-3
\sqrt{201}}}\right)$
$c=3$
Solve[Log[1/2 (-E^x + Sqrt[2 - 3 E^(2 x)])] == 3 x, x]
$x=-\frac{1}{2} \log \left(\frac{6}{-2-\frac{4\ 2^{2/3}}{\sqrt[3]{41+3 \sqrt{201}}}+\sqrt[3]{2 \left(41+3 \sqrt{201}\right)}}\right)$
Numeric solution
You can define an inverse function
x = InverseFunction[Log[1/2 (-E^# + Sqrt[2 - 3 E^(2 #)])]/# &]
that would still yield exact solutions when it can, for instance for $c=2$:
x[2] // ToRadicals // Simplify
$\log \left(-\frac{1}{4}+\frac{1}{4 \sqrt{\frac{3}{-5-\frac{20\ 2^{2/3}}{\sqrt[3]{49+3 \sqrt{489}}}+2 \sqrt[3]{98+6 \sqrt{489}}}}}+\frac{1}{2}
\sqrt{-\frac{5}{6}+\frac{5\ 2^{2/3}}{3 \sqrt[3]{49+3 \sqrt{489}}}-\frac{\sqrt[3]{49+3 \sqrt{489}}}{3\ 2^{2/3}}+\frac{3}{2} \sqrt{\frac{3}{-5-\frac{20\
2^{2/3}}{\sqrt[3]{49+3 \sqrt{489}}}+2 \sqrt[3]{98+6 \sqrt{489}}}}}\right)$
but otherwise works numerically:
In[]:= N[x[2]]==x[2.]
Out[]= True
and even plots:
Plot[x[c], {c, 0, 7}, PlotRange -> All, PlotTheme -> "Detailed"]

s[c_] := 1/2 Log[Root[-1 + 2 #1 + 2 #1^((1 + c)/2) + 2 #1^c &, 1]]
$\endgroup$