How do I solve this exponential equation? $$x = 2^{x-3}$$
Does there exist an analytical solution to this equation? If so, how do I find it?
What if it is changed to an equality?
$$x>2^{x-3}$$
 A: 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 $(x_1,x_2)$
A: The equality $x = 2^{x-3} = \text{e}^{x \ln 2}\, 2^{-3}$ rewrites
$$ -x \ln 2 \,\text{e}^{-x \ln 2} = -2^{-3} \ln 2\, , $$
which is of the form $X\text{e}^{X} = Y$ with $X=-x \ln 2$ and $Y=-2^{-3} \ln 2$. The solutions are $X=W(Y)$, where $W$ is the Lambert W-function. Therefore,
$$ x = -\frac{W(-2^{-3} \ln 2)}{\ln 2} \quad\in\quad \lbrace \underbrace{0.1375\dots}_{x_1}, \underbrace{5.4449\dots}_{x_2}\rbrace\, . $$
Both functions $x\mapsto x$ and $x\mapsto 2^{x-3}$ are continuous. Since $0 < 2^{0-3}$ and $6 < 2^{6-3}$, the inequality $x>2^{x-3}$ is necessarily satisfied over $\left]x_1,x_2\right[$.
