# How to solve exponential function inequality?

How do I solve the exponential equations like $2^\frac{x}{8}<x$? I can solve this by plotting into graph. But is there any way to do it mathematically? or like $2^x < 100x^2$ . I am trying to actually solve the thing. ie for which set of values the inequality holds...

• This is somewhat implicit in the discussion of the Lambert $W$ function but it's worth pulling out explicitly: if your $x$ is not a rational number (e.g., the solutions of $2^x=2x$) then it can't be algebraic, as a consequence of the Gelfond-Schneider theorem ( en.wikipedia.org/wiki/Gelfond–Schneider_theorem ). Since you can eliminate non-integer rationals fairly easy in your sample cases too (e.g. because $2^k$ is only rational for integer $k$) this means that any solutions must be transcendental. – Steven Stadnicki Feb 28 '14 at 16:42

## 3 Answers

Let's examine this WolframAlpha output: From there, it's easy to see geometrically what the solution set looks like. (It sounds like you knew this much.) We also see the solution set expressed in terms of a mysterious function $W$, which is a special function called the Lambert $W$ function. You can click the approximate form button to find that (approximately) $1.1<x<44.559$ but I don't believe that the solution set can be expressed in a simpler form.

• Can you please give me the link of the wolfram page where you inputted the graph? – Tamim Addari Feb 28 '14 at 14:00
• I believe it's there where I typed WolframAlpha output. Oh hey, there it is again!! – Mark McClure Feb 28 '14 at 14:02

For the first question, the problem is basically to solve the equation $$f(x)=2^{x/8}- x=0$$ The solutions are obtained using Lambert function and the solution are $$x_1=-\frac{8 W\left(-\frac{\log (2)}{8}\right)}{\log (2)}$$ $$x_2=-\frac{8 W_{-1}\left(-\frac{\log (2)}{8}\right)}{\log (2)}$$ which are respectively equal to $1.10$ and $43.56$. So, the inequality is satisfied between these two roots.

In practice, any equation of the form $$a+b x+c \log (d+e x) =0$$ has solutions expressed using Lambert function.

The same happens with the second equation you gave; the three roots of the equations are also given in terms of Lambert function; they correspond to $x_1=-0.10$, $x_2=0.10$ and $x_3=14.32$.

Your equations can't be solved algebraically (without involving Lambert's $W$), but numerical instead.