Answer: D

Small Hint:

Compare $1+x$ with an exponential function for $x\gt0$

Big Hint:

The standard inequality $\ln(1+x)\lt x$ is even stronger for common logarithms

Solution:

For $x\gt0,$ the standard exponential inequality gives $1+x\lt e^x.$ Taking natural logarithms yields $$\ln(1+x)\lt x.$$ If $\log$ denotes the common logarithm, then $$\begin{aligned} \log(1+x)&=\frac{\ln(1+x)}{\ln10}\\ &\lt\ln(1+x), \end{aligned}$$ so the same listed inequality holds.

Thus, the correct answer is D.