Acceleration as the logarithmic derivative of the redshift factor $U^\sigma \nabla_\sigma U^\mu = g^{\mu\nu}\nabla_\nu \ln V$ Let $K^\mu = V(x)U^\mu$ be vector proportional to the four-velocity $U^\mu$. (e.g. $K^\mu$ is a normalized time-like Killing vector for an observer at infinity).
Then, $V(x) = \sqrt{-K_\nu K^\nu}$ is called the "redshift factor". 
Also, the four-acceleration $a^\mu$ is given by $U^\sigma \nabla_\sigma U^\mu$.
According to Sean Carroll, Spacetime and Geometry, p. 247, $a_\mu = \nabla_\mu\ln V$. Why?
Attempt:
\begin{align}\nabla_\mu\ln V &= \frac 1{2V^2}\nabla_\mu\left(-K_\nu K^\nu\right)\\
&=-\frac 1{2V^2}\left((\nabla_\mu K_\nu)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\
&=-\frac 1{2V^2}\left((\nabla_\mu g_{\rho\nu}K^\rho)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\
&=-\frac 1{2V^2}\left(K_\rho\nabla_\mu K^\rho + K_\nu\nabla_\mu K^\nu\right)\\
&=-\frac {K_\nu\nabla_\mu K^\nu}{V^2}\\
&= -\frac 1{V}U_\nu\nabla_\mu\left(VU^\nu\right)\\
&= -U_\nu\nabla_\mu U^\nu - \frac 1 VU_\nu U^\nu\nabla_\mu V\end{align}
 A: At some point in your calculation, you got the following:
$$\nabla_\mu \ln{V} =-\frac{1}{V^2} K_\nu \nabla_\mu K^\nu$$
Using the metric compatibility, we can write
$$\nabla_\mu \ln{V} =-\frac{1}{V^2} K_\nu \nabla_\mu \left( g^{\nu \lambda} K_\lambda \right)=-\frac{1}{V^2} K_\nu g^{\nu \lambda}  \nabla_\mu K_\lambda=-\frac{1}{V^2} K^\lambda \nabla_\mu K_\lambda.$$
Now, since $K^\mu$ is a Killing vector field, then it satisfies the Killing equation:
$$ \nabla_\mu K_\lambda + \nabla_\lambda K_\mu =0.$$
Substituting in the equation above, we get
$$\nabla_\mu \ln{V}= \frac{1}{V^2} K^\lambda \nabla_\lambda K_\mu.$$
Using the relation $K^\lambda= V(x) U^\lambda$, we get
$$\nabla_\mu \ln{V}= \frac{1}{V^2} \left( VU^\lambda \right) \nabla_\lambda \left( V U_\mu \right).$$
Applying the Leibniz rule,
$$\nabla_\mu \ln{V}= \frac{1}{V} U^\lambda \left( (\nabla_\lambda V) U_\mu + (\nabla_\lambda U_\mu) V \right).$$
$$\therefore \nabla_\mu \ln{V}= U^\lambda U_\mu  \frac{1}{V} \nabla_\lambda V  + U^\lambda \nabla_\lambda U_\mu$$
Or, using $a_\mu = U^\lambda \nabla_\lambda U_\mu$ and $\frac{1}{V} \nabla_\lambda V =\nabla_\lambda \ln{V}$,
$$ \nabla_\mu \ln{V}= U_\mu U^\lambda \nabla_\lambda \ln{V}  + a_\mu   \ \ \ \ \ \rightarrow   (1)$$
Now, contract both side of equation (1) with $U^\mu$ and use the fact that $a_\mu U^\mu=0$ and $U^\mu U_\mu=-1$.
$$\therefore U^\mu \nabla_\mu \ln{V}= - U^\lambda \nabla_\lambda \ln{V}.$$
The only way this is satisfied is to have
$$U^\lambda \nabla_\lambda \ln{V}=0 \ \ \ \ \ \rightarrow (2)$$
Substitute from equation (1) into equation (2), you get:
$$ \nabla_\mu \ln{V}=a_\mu,$$
which is what we wanted to prove.
