Let $Z$ be a positive semidefinite matrix with nonnegative entries, and define $X=\log(1+Z)$, where the $\log$ is taken entrywise, i.e., $X_{ij}=\log(1+Z_{ij})$. Are there some simple sufficient conditions that guarantee that $X$ is positive semidefinite?

**Note 1**
Intuitively, this should work when $Z$ is small enough, because then $X \simeq Z \succeq 0$.

**Note 2** A simple counter-example is given by $Z=\left[ \begin{array}{cc}
3 & 5\\
5 & 9
\end{array}
\right]$.

**Note 3**
It is easy to verify that the converse relation allways holds, that is,
$$X\succeq 0 \Longrightarrow Z:=exp(X)-1\succeq 0,$$
where the above operations are elementwise: $Z_{ij} = e^{X_{ij}} -1$. Indeed, we have the relation $Z=\sum_{k=1}^\infty X^{\circ k}/k!$, where $X^{\circ k}$ is the $k$th power of $X$ with respect to the Hadamard (elementwise) product, and it is well-known that the Hadamard product of 2 positive semidefinite matrices is positive semidefinite.

**Note 4** Some background on this question: I ask myself when a random variable $Y$ with mean vector $\mu\in\mathbb{R}^n$ and variance-covariance matrix $\Sigma \succeq 0 \in \mathbb{R}^{n\times n}$ can be fitted to a multivariate log-normal distribution using the method of moments. Applying the formulas, we find that the elements of variance-covariance matrix $V$ of $\log Y$ must satisfy
$$V_{ij}= \log(\frac{\Sigma_{ij}}{\mu_i \mu_j} +1),$$
but in general this matrix is not guaranteed to be positive semidefinite.