$\sqrt{2^2} = 2$, $\sqrt{2^4} = 2^2 = 4$, $\sqrt{2^6} = 8 \neq 6$, $\ldots$

Question

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 $s \in \mathbb{N}$.

Answer 1

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.

Answer 2

my proof using differentiation

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

$$f(2)=2^2-2×2=4-4=0 \cdots (1)$$

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

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

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

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

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

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

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

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

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

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

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

Answer 3

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 $

$2×2t =4t \gt 2(t+1)$.