Family of curves that transitions smoth from linear to logarithmic I am looking for a family of curves that smoothly transitions from
$f(x)=x$
to
$f(x)=log(x+1)$
 A: In general, you can combine any two functions $f_0(x)$ and $f_1(x)$ using a third (interpolating) function $w(y)$:
$$f(x, y) = \biggl(1 - w(y)\biggr) f_0(x) + w(y) f_1(x) \tag{1a}\label{1a}$$
such that
$$\begin{aligned}
f(x, 0) &= f_0(x) \\
f(x, 1) &= f_1(x) \\
\end{aligned}$$
and $f(x, y)$ for $0 \lt y \lt 1$ are "combinations" of the two functions.  The interpolating function $w(y)$ defines how the two functions get mixed.
Any function that fulfills $w(0) = 0$, $w(1) = 1$, $0 \lt w(y) \lt 1$ for $0 \lt y \lt 1$ will work, but some common (symmetric) ones are
$$\begin{aligned}
w_1(y) &= y \\
w_3(y) &= 3 y^2 - 2 y^3 \\
w_5(y) &= 10 y^3 - 15 y^4 + 6 y^5 \\
w_\lambda(y) &= \begin{cases}
0, & y \le 0 \\
2^{\lambda - 1} y^\lambda, & 0 \lt y \le \frac{1}{2} \\
1 - 2^{\lambda - 1} (1 - y)^\lambda, & \frac{1}{2} \lt y \lt 1 \\
1, & y \ge 1 \\
\end{cases} \end{aligned} \tag{1c}\label{1c}$$
If you want replacing $y$ with $1-y$ to behave as if one swapped $f_0$ and $f_1$, then you do want a symmetric function, one that for $0 \le y \le 1$ fulfills $w(y) + w(1-y) = 1$.
Note that outside $0 \le y \le 1$, we actually do extrapolation, which may or may not make any sense at all.  Again, it depends on the behaviour of the intepolating/extrapolating function $w(y)$.

For example, using linear interpolation, $w(y) = y$, and $f_0(x) = x$, $f_1(x) = \log(x + 1)$, we get
$$f(x, y) = (1 - y) x + y \log(x + 1)$$
and $f(x, 0) = x$, $f(x, 1) = \log(x + 1)$, and something in between for $0 \lt y \lt 1$ using $f(x, y)$.
