Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.1 Junior Find the answer

If $x>0$ , then the correct relationship is:
$\textbf{(A)}\ \log (1+x) = \frac{x}{1+x}$ $\textbf{(B)}\ \log (1+x) x$ $\textbf{(D)}\ \log (1+x) < x$ $\textbf{(E)}\ \text{none of these}$

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is D, and the solution below works it through.

Solution

Plug in $x=9$ . Then, you can see that the answer is $\fbox{D}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.