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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...

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    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Feb 28 '14 at 16:42
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Let's examine this WolframAlpha output:

enter image description here

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.

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  • $\begingroup$ Can you please give me the link of the wolfram page where you inputted the graph? $\endgroup$ – Tamim Addari Feb 28 '14 at 14:00
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    $\begingroup$ I believe it's there where I typed WolframAlpha output. Oh hey, there it is again!! $\endgroup$ – Mark McClure Feb 28 '14 at 14:02
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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 $x_3=14.32$.

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

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