I have currently found a formula
...view formula here
$$y = 1 - (1-x^u)^{1/u}$$ $$\quad u =\frac{v}{2(1 - v)}$$ $$\quad v = t^{\frac{\ln 1.5}{\ln 2}}$$
which for t values from 0 to 1,
gives a very smooth curve on x-axis between 0 and 1.
however as you can see that it is extremely hard to calculate for any practical purposes for rendering because of exponentials.
can anyone find simpler one or make it small?
Edit: Removing comment off-topic
You can express the formula as follows, $$y = 1 - (1-x^u)^{1/u}, \quad u =\frac{v}{2(1 - v)}, \quad v = t^{\frac{\ln 1.5}{\ln 2}}$$ Maybe this breakdown will help? At least it can reduce repeated calculations. Since $x$ and $t$ are small, you could try Taylor series expansions with a few terms.