$\sqrt{2^2} = 2$, $\sqrt{2^4} = 2^2 = 4$, $\sqrt{2^6} = 8 \neq 6$, $\ldots$ In general, if $t \in \mathbb{N}$, when is it true that
$$\sqrt{2^{2t}} = 2^t = 2t?$$
I know of course that it is true when $t - 1 = \log_{2}(t)$.  Are there other (more arithmetic) conditions, for which the equation $2^t = 2t$ holds, for $t \in \mathbb{N}$?
It appears to be true when $t = 2^s$, for some $s \in \mathbb{N}$.
 A: The first equality is true: $\sqrt{2^{2t}}=\sqrt{2^{t+t}}=\sqrt{2^t\cdot2^t}=2^t$.
The second holds only for $t=1$ and $t=2$. Indeed, define for $t\in\Bbb R$ $$f(t)=\cfrac{2^{t}}{2t}$$ and $$g(t)=\ln f(t)=t\ln 2-\ln 2-\ln t$$
Now,
$$g'(t)=\ln2-\frac1t$$
We see that $g'$ vanishes only at one point, namely $t=1/\ln 2$. By Rolle's theorem, $g$ has no more than two zeros, and hence the equation $f(t)=1$ can not have more than two solutions.
A: Proposition:
$2^t \gt 2t$ for $t \gt 2,$  $ t \in \mathbb{N}.$
Proof by induction:
0)True for $t=3.$
1) Assume true for $t.$
2) Step: Show for $t+1.$
$2×2^t = 2^{t+1} \gt $
$2×2t =4t \gt 2(t+1)$.
A: 
my proof using differentiation

Let
$$f(t)=2^t-2t$$

$$f(2)=2^2-2×2=4-4=0 \cdots  (1)$$

$$f'(x)=\ln(2) 2^t  -2$$

$$f'(2)=4\ln(2) -2=0.772589\cdots(2) $$

$$f''(t)=(\ln(2))^2 2^t>0$$
$$\text{for }x\in R$$

$$\text{.: f'(t) is an increasing function for }x\in R$$

$$\text{but }f'(2)>0$$

$$\text{.:} f'(t)>0 \text{ for }x\gt2$$

$$\text{.: }f(t)\text{ is an increasing function for }x\gt 2$$

$$\text{but }f(2)=0$$

$$\text{.: }f(2)\gt 0 \text{ for }x\gt2$$

$$\text{.:}2^t-2t \gt 0 \text{ for }x\gt2$$

$$\text{.:}2^t\gt 2t\text{ for }x\gt 2$$
