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

1. Domain of the Logarithm

For a real logarithm, the inside must be positive: $1+x>0 \;\; \Longrightarrow \;\; x>-1$

2. Turn the Inequality into a Friendlier Function

Move everything to one side: $\log(1+x) \le x \quad\Longleftrightarrow\quad x-\log(1+x) \ge 0$ Define $g(x)=x-\log(1+x)$

3. Analyze the Shape with Derivatives

Differentiate to see where $g(x)$ rises or falls: $g'(x)=1-\frac{1}{1+x}=\frac{x}{1+x}$

• If $-1<x<0$, then $x<0 \Rightarrow g'(x)<0$ → decreasing.
• If $x>0$, then $x>0 \Rightarrow g'(x)>0$ → increasing.

Thus $g(x)$ has a minimum at $x=0$.

4. Evaluate the Minimum Point

$g(0)=0-\log(1)=0$ Because $g(x)$ is never below this minimum, we get $g(x)\ge0$ for every $x>-1$. That translates back to $\log(1+x)\le x \quad \text{for all}\; x>-1$

5. Final Interval

Combine with the domain: the complete solution set is $(-1,\infty)$

Imagine Climbing a Hill

• Think of the hill as the graph of two paths:
  1. The log path that shows how high you get with $\log(1+x)$.
  2. The straight path that simply goes up with $x$.
• We want to check where the log path never rises above the straight path (that is, $\log(1+x) \le x$).
• First rule of the log: you’re only allowed to use it when the inside is positive, so 1 + x must be more than 0. That means $x>-1$.
• At $x = 0$, both paths meet at the same spot ($\log(1)=0$).
• Any step left of 0 (but still more than -1) makes the log path drop faster than the straight path, so the log path stays below.
• Any step right of 0 makes the straight path climb faster, so again the log path stays below.
• Conclusion: Every place you can legally stand ($x>-1$) works! The only forbidden spot is $x\le-1$.

So the answer is all numbers bigger than -1.

Step 1: State Domain

$1+x>0 \;\Rightarrow\; x>-1$

Step 2: Rewrite Inequality

$\log(1+x) \le x \;\Longleftrightarrow\; x-\log(1+x) \ge 0$ Define $g(x)=x-\log(1+x)$

Step 3: Differentiate

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

Step 4: Monotonicity

• For $-1<x<0$, $g'(x)<0$ → $g$ decreases.
• For $x>0$, $g'(x)>0$ → $g$ increases. Hence $x=0$ is the minimum.

Step 5: Evaluate Minimum

$g(0)=0-\log(1)=0$ Since the minimum is $0$ and $g(x)\ge0$ everywhere else, the inequality holds for the whole domain.

Step 6: Combine with Domain

$\boxed{(-1,\infty)}$ (Equality only at $x=0$.)