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$$x = 2^{x-3}$$

Does there exist an analytical solution to this equation? If so, how do I find it?

What if it is changed to an equality? $$x>2^{x-3}$$

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You may need the Lambert-W function. –  sos440 Nov 3 '12 at 3:29
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1 Answer

Consider $f(x) = 2^{x-3} - x$. We then have $f(3) < 0$ and $f(6), f(0) > 0$.

$f'(x) = \dfrac{2^x \log_e(2)}8 - 1> 0$, for $2^x > \dfrac{8}{\log_e(2)}$ i.e. for $x > 3 - \log_2( \log_e(2))$

For $x>4$, the function is increasing and for $x<3$, the function is decreasing. Hence there are only two roots, one ($x_1$) in the vicinity of $0$ and the other ($x_2$) in the vicinity of $5$.

For the inequality, we are interested in the region where $f(x) < 0$, which is nothing but the region between the two roots i.e in the interval $(x_1,x_2)$

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What about the inequality? Is there a way to solve that with an exact answer? –  sdf Nov 3 '12 at 3:44
    
The inequality is true for x=1 and false for $x \ge 6$. I'm not sure what you mean by "solve that with an exact answer". –  marty cohen Nov 3 '12 at 6:10
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