If 2^x=x+2. solve it and prove x=2.
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Define $f(x)=2^x-x-2$.
$f'(x)=2^x\ln 2-1$
Find critical points:
$x=\dfrac{-\ln (\ln 2)}{\ln 2} \approx 0.5288$. At $x=0, f'(x)<0$ and at $x=1$, $f'(x)>0$. Since $2^0-0-1 <0$, to see if there is another solution, we take the limit as $x\to -\infty $.
$\displaystyle \lim_{x \to -\infty}(2^x-x-2)$ does not exist, as it tends towards infinity. This indicates that there exists exactly one $x <0$ such that $f (x)=0$. While I am able to prove the existence of another solution, I am not sure what it is exactly. It is close to $x\approx -1.69009$