Solving $ 1.2^x \le \log_{1.2} (x) $ I am stumped on this question, the best I can do is find a numerical solution but I would prefer to find an exact solution.

The question is to solve:
$$ 1.2^x \le \log_{1.2} (x) $$

According to Wolfram the numerical, or approximate solution is $0<x\le1.25773$ and $ x\ge14.7675$.
My attempt:
$$ 1.2^x \le \log_{1.2} (x) 
\\ ~~~~~~~~~~~~ x \ge 1.2^{1.2^x} ~, ~~\text{ for } x >0 
$$
Alternatively $ x \le \log_{1.2} \left( \log_{1.2} (x) \right)   ~,~~ x >0 $.
The thorniness is that $x$ is stuck on both sides of the inequality. Also the intended audience of the question is a 10-th grade Maths Class page 9, number 68.
Perhaps the exact solution is something that becomes obvious in hindsight, like turning $1.2$ into $\frac 6 5$, and use properties of exponents.
We could attempt to solve $ 1.2^x = \log_{1.2}(x) $ or equivalently $ x= 1.2^{1.2^x} $ , since graphically the intersection of $f(x)=1.2^x $ and $g(x) = \log_{1.2}(x)$ determine the intervals where the inequality is true.
 A: See the important Edit I wrote at the bottom of this answer.

Fig. 1: On this (Desmos) graphical representation, one retrieves the intervals found with Wolfram Alpha. More precisely, one "sees" that the exponential curve is under the logarithmic curve for $x \in [a , b]$ which is the solution of your issue. A fully rigorous answer should invoke the convexity (resp. concavity) of $\exp_{1.2}$ (resp $\log_{1.2}$).
These two functions are mutually inverse ; as such, their graphical representations are symmetrical with respect to the line with equation $y=x$ with crossing points (limits of intervals of validity for the inequation) given by the solutions of equations :
$$1.2^x = \log_{1.2}(x) \color{red}{ = x }$$
What you have to solve now is either equation:
$$1.2^x= x \tag{1}$$
or, equivalently $$ \log_{1.2}(x) = x \tag{2}$$
Solving (1) or (2) is indeed simpler than solving:
$$1.2^x = \log_{1.2}(x) \tag{3}$$
A first answer is indeed using a natural numerical method called "fixed point iteration" with this recurrently defined sequence
$$u_{n+1}=1.2^{u_n} \ \text{with (for example)} \ u_0=1$$
It converges towards the abscissa $a \approx 1.26$ of the intersection point closer to the origin, and  sequence
$$v_{n+1}=\log_{1.2}(v_n) \ \text{with (for example)} \ v_0=10$$
which will converge to the abscissa $b \approx 14.8$ of the intersection point farthest from the origin.

Edit: As your question is mainly about the existence of closed form formulas for $a$ and $b$, here is an answer using Lambert W function. Let us begin by $a$:
$$a=W(\alpha)/\alpha \ \ \ \text{where} \ \ \ \alpha:=-\ln(1.2) \tag{5}$$
(more precisely, we have used the principal branch $W_0$ of this function : this precision will be important for the second root).
I checked the exactness: Wolfram Alpha gives:
$$a=1.2577345413765...$$
Explanation of relationship (5): (1) is equivalent to
$$-\ln(1.2)=-\ln(1.2)xe^{-\ln(1.2)x}$$ which can be written
$$-\ln(1.2)=w e^{w} \ \iff \ w=W(-\ln(1.2)) \ \text{with} \ w:=-\ln(1.2)x$$
The second root $b$ can be expressed under a similar closed form
$$b=W_{\color{red}{-1}}(\alpha)/\alpha=14.76745838...\tag{6}$$
but now with $W_{-1}$, which is a different branch of Lambert $W$. In this case, the Wolfram Alpha computation request is like this.
Remark: expression found in (5) can be expressed under the form of an infinite "tetration"
$$W(x)/x=x^{x^{x^x ...}}$$
(see formula (3) of the reference given above).
