The following data are inferred from a presentation slide, so I do not much info.
Using linear approximation and log rules $\sqrt x $ can be rewritten as $\frac{x+1}{2}$, where $(1 \leq x \lt 2) $ ... (1)
I understood the Eq. (1).
Now the slide says, in the range $(2 \leq x \lt 4)$,
$ \sqrt x = 2^{\frac{log_2(x)}{2}} \approx 2^{\frac{(1+x-1)}{2}}$ ... (2)
Could some one tell me, how and why do we get the term `$x$' approximated to $ (1+x-1) $ in Eq. (2) ?
Note! I am aware that using linear approximation $log_2(x) \approx x-1 $
Thank You.
Probably there is a typo in $ \sqrt x \approx 2^{\frac{(1+x-1)}{2}}$
because it is not a correct approximate in the range $2 \leq x < 4$
Case of $x=2 $ then $ \sqrt 2 \approx 2^{\frac{(1+2-1)}{2}}=2$ is false.
Case of $x=4 $ then $ \sqrt 4 \approx 2^{\frac{(1+4-1)}{2}}=4$ is false.
If we use the Taylor's series : $$f(x)=f(x_1)+f'(x_1)(x-x_1)+...$$ with $x_1=2$ and $f(x)=\sqrt{x}$ hense $f'(x)=\frac{1}{2\sqrt{x}}$ we obtain :
$$f(x)\approx \sqrt{2}+\frac{1}{2\sqrt{2}}(x-2)$$
Case of $x=2 $ then $ \sqrt 2 \approx \sqrt{2}+\frac{1}{2\sqrt{2}}(2-2)=\sqrt{2}$ is OK.
Case of $x=4 $ then $ \sqrt 4 \approx \sqrt{2}+\frac{1}{2\sqrt{2}}(4-2)=\frac{3\sqrt{2}}{2} \simeq 2.12$ is OK.
Another kind of approximate on the form $\sqrt{x} \approx 2^{ax+b}$ is identifed so that $2^{2a+b}=\sqrt{2}$ and $ 2^{4a+b}=2$ leading to $a=\frac{1}{4}$ and $b=0$ : $$\sqrt{x}\approx 2^{x/4}$$ on the range$2 \leq x < 4$
but the preceeding approximate $f(x)\approx \sqrt{2}+\frac{1}{2\sqrt{2}}(x-2)$ is better.
Of course, even better approximates can be obtained in considering more terms of the Taylor's series.