The set of all $x$ for which $\log (1+x) \leq x$, is

$(-1, \infty)$

Let $f(x)=\log (1+x)-x$

Clearly, f(x) is defined for all x > -1.

Now,

$f(x)=\log (1+x)-x$

$\Rightarrow f'(x)=\frac{1}{1+x}-1=-\frac{x}{1+x}$

$\Rightarrow f'(x)<0$ for $x>0$ and $f'(x)>0$ for $-1<x<0$

$\Rightarrow f(x) \leq f(0)$ for $-1<x<\infty$

$\Rightarrow \log (1+x)-x \leq 0$ for $-1<x<\infty$

$\Rightarrow \log (1+x) \leq x$ for $x \in(-1, \infty)$