Solving $a_{n+1} = x^{a_n}$ for various $x$ 
(a) Consider a sequence defined by $a_0=1$ and $a_{n+1}=(\sqrt2)^{a_n}$ . Prove that limit exists and find it.


(b) Show that the limit doesn't exist finitely if we replace $\sqrt2$ by $1.5$. What are the maximum and minimum numbers such that limit exists?

I could do the first question and the first part of the second question.
For (a), by induction, each term of the sequence lies in $(1,2)$ and is therefore monotonically increasing by the recursion function. Hence the limit exists, let it be $l$, it comes $2$.
For (b), the sequence is monotone unbounded above, and the limit is $\infty$ similarly.
How do I do the second part of (b)?

From my comment, we want to find when a tangent intersection of the curves $t$ and $x^t$ will occur. This means we have to satisfy the following system of equations:
$$\begin{cases}t = x^t \\ 1 = \log x \cdot x^t\\ \end{cases}$$
with the added constraint that we are looking for the solution $x > 1$
