Solving for terms of the sequence given by $ x^{2(2^k-1)}(x-1)=\ln2$ or gain some insight I'm interested in properties of the sequence given by (int k) :
$$ x^{2(2^k-1)}(x-1)= \ln2$$
$k = 1$: $x^2(x-1) = \ln2$     or  $x^3-x^2 = \ln2$  :$x \approx 1.3737...$
$k=2$: $x^6(x - 1) = \ln2$   or   $x^7 - x^6 = \ln2$ : $x\approx1.2175...$
$k=3$: $x^{14}(x-1) = \ln2$   or   $x^{15} - x^{14} = \ln2$: $x\approx1.1284...$
Obviously this is asking for the base when certain  consecutive powers differ by $\ln2$. Also apparent is how it approaches 1.
Unfortunately, I'm not able to solve them generally. Might there be some recursive property from one element to the next. Or maybe some series representation. Just looking for some insight and/or pointed in a good direction for what book I should crack open. Thanks in advance. ps. I'm just a novice hobbyist

One can make approximations of the solution building, around $x=1$, the $[1,n]$ Padé approximant of
$$x^{2(2^k-1)}(x-1)- \log(2)$$
To fit in the page, I limited to $n=4$ and $a=\log(2)$. The approximation is
$$x=1+\frac{4 a^4 \left(2^k-2\right) \left(2^k-1\right) \left(2^{k+1}-3\right)+12 a^3
   \left(-9\ 2^k+4^{k+1}+5\right)+36 a^2 \left(2^k-1\right)+6 a}{a
   \left(2^k-1\right) \left(a \left(a \left(2^{k+1}-3\right) \left(a
   \left(2^k-2\right) \left(2^{k+1}-5\right)+2^{k+5}-40\right)+18 \left(3\
   2^{k+1}-7\right)\right)+48\right)+6}$$
This would give the following results
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
 1 & 1.36953 &  1.36955 \\
 2 & 1.21408 &  1.21522 \\
 3 & 1.12998 &  1.12815 \\
 4 & 1.07960 &  1.07632 \\
 5 & 1.04724 &  1.04507 \\
 6 & 1.02664 &  1.02630 \\
 7 & 1.01434 &  1.01516 \\
 8 & 1.00747 &  1.00864 \\
 9 & 1.00382 &  1.00486 \\
 10 & 1.00193 &  1.00271
\end{array}
\right)$$
and I suppose that using this extimate as $x_0$, Newton method should converge in a couple of iterations.
