Solving $ 1.2^x \le \log_{1.2} (x) $

Solution 1:

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).