Mean of the log and variance of the log I am struggling to derive the following identities:
$$
\mathbb{E}[\log Z]=2\log(\mathbb{E}[Z])-\frac12\log(\mathbb{E}[Z^2])
$$
$$
\mathrm{Var}[\log Z]=\log(\mathbb{E}[Z^2])-2\log(\mathbb{E}[Z])
$$
where $Z$ is a random variable. These are equations (B6) and (B7) in this paper.

$\newcommand{\e}{\operatorname E} \newcommand{\v}{\operatorname{var}}$The paper actually assumes $X= \log\mathcal Z$ is normally distributed, not just that it is some random variable.
Let $\mu=\e(X) = \e(\log\mathcal Z)$ and $\sigma^2 = \v(X) = \v(\log\mathcal  Z).$
Then $\mathcal Z=e^X = \exp(X)$ and
\begin{align}
\e(\mathcal Z) = {} &\int_{-\infty}^{+\infty} (\exp x) \varphi_{\mu,\sigma^2}(x) \,dx \\
& \text{where } \varphi_{\mu,\sigma^2} \text{ is the normal} \\
& \text{density with expectation $\mu$} \\
& \text{and variance $\sigma^2$.} \\[8pt]
= {} & \int_{-\infty}^{+\infty} (\exp x) \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12 \left( \frac{x-\mu} \sigma \right)^2 \right) \,\,\frac{dx}{\sigma} \\[8pt]
= {} & \int_{-\infty}^{+\infty} (\exp(\mu + \sigma w)) \frac 1 {\sqrt{2\pi}} \exp \left( -\frac 12\,w^2 \right)\, dw \\[8pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12 w^2 + \sigma w + \mu \right) \, dw \\[8pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12\left( w^2 - 2\sigma w \right) + \mu \right) \, dw \\[8pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12\left( w^2 - 2\sigma w + \sigma^2 \right) + \mu + \frac12\sigma^2 \right) \, dw \\
& \text{This is completing the square.} \\[8pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12(w-\sigma)^2 \right) \underbrace{ \exp\left( \mu + \frac12\sigma^2 \right) }_\text{No “$w$” appears here!} \, dw \\[8pt]
& \text{The absence of $w$ from the expression above} \\
& \text{the $\underbrace{\text{underbrace}}$ means that that can be pulled out:} \\[8pt]
= {} & \exp\left( \mu + \frac12\sigma^2 \right)  \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12(w-\sigma)^2 \right) \, dw \\[8pt]
= {} & \exp\left( \mu + \frac12\sigma^2 \right) \cdot 1
\end{align}
That last integral is equal to $1$ because it is the integral of the normal density with expectation $\sigma$ and variance $1.$
We will also need $\e(\mathcal Z^2).$ Notice that
$$
\Big( \exp(\mu+\sigma w) \Big)^2 = \exp(2\mu + 2\sigma w).
$$
So the random variable $\mathcal Z^2= \exp(2X)$ has values of $\mu$ and $\sigma$ that are twice as big; hence we have
$$
\e(\mathcal Z^2) = \exp\left(2\mu + \frac12(2\sigma)^2 \right).
$$
So
\begin{align}
& \log\e(\mathcal Z) = \mu + \frac12 \sigma^2, \\[8pt]
& \log\e(\mathcal Z^2) = 2\mu + \frac12(2\sigma^2) = 2\mu + 2\sigma^2, \\[8pt]
\text{and so } \mu & = \e(\log\mathcal Z) = 2\log\e(\mathcal Z) - \frac12 \log\e(\mathcal Z^2) \\[8pt] 
\sigma^2 & = \v( \log\mathcal Z) = \log\e(\mathcal Z^2) - 2\log\e(\mathcal Z).
\end{align}
