Interpretation of log transformed predictor and/or response I'm wondering if it makes a difference in interpretation whether only the dependent, both the dependent and independent, or only the independent variables are log transformed.
Consider the case of
log(DV) = Intercept + B1*IV + Error 

I can interpret the IV as the percent increase but how does this change when I have 
log(DV) = Intercept + B1*log(IV) + Error

or when I have 
DV = Intercept + B1*log(IV) + Error

?
 A: Charlie provides a nice, correct explanation.  The Statistical Computing site at UCLA has some further examples:
https://stats.oarc.ucla.edu/sas/faq/how-can-i-interpret-log-transformed-variables-in-terms-of-percent-change-in-linear-regression, and
https://stats.oarc.ucla.edu/other/mult-pkg/faq/general/faqhow-do-i-interpret-a-regression-model-when-some-variables-are-log-transformed
Just to complement Charlie's answer, below are specific interpretations of your examples. As always, coefficient interpretations assume that you can defend your model, that the regression diagnostics are satisfactory, and that the data are from a valid study.
Example A: No transformations
DV = Intercept + B1 * IV + Error 

"One unit increase in IV is associated with a (B1) unit increase in DV."
Example B: Outcome transformed
log(DV) = Intercept + B1 * IV + Error 

"One unit increase in IV is associated with a (B1 * 100) percent increase in DV."
Example C: Exposure transformed
DV = Intercept + B1 * log(IV) + Error 

"One percent increase in IV is associated with a (B1 / 100) unit increase in DV."
Example D: Outcome transformed and exposure transformed
log(DV) = Intercept + B1 * log(IV) + Error 

"One percent increase in IV is associated with a (B1) percent increase in DV."
A: In the log-log- model, see that
$$\begin{equation*}\beta_1 = \frac{\partial \log(y)}{\partial \log(x)}.\end{equation*}$$
Recall that
$$\begin{equation*} \frac{\partial \log(y)}{\partial y} = \frac{1}{y} \end{equation*}$$
or
$$\begin{equation*} \partial \log(y) = \frac{\partial y}{y}. \end{equation*}$$
Multiplying this latter formulation by 100 gives the percent change in $y$. We have analogous results for $x$.
Using this fact, we can interpret $\beta_1$ as the percent change in $y$ for a 1 percent change in $x$.
Following the same logic, for the level-log model, we have
$$\begin{equation*}\beta_1 = \frac{\partial y}{\partial \log(x)} = 100 \frac{\partial y}{100 \times \partial \log(x)}.\end{equation*}$$
or $\beta_1/100$ is the unit change in $y$ for a one percent change in $x$.
A: The main purpose of linear regression is to estimate a mean difference of outcomes comparing adjacent levels of a regressor. There are many types of means. We are most familiar with the arithmetic mean. 
$$AM(X) = \frac{\left( X_1 + X_2 + \ldots + X_n \right)}{n}$$
The AM is what is estimated using OLS and untransformed variables. The geometric mean is different:
$$GM(X) = \sqrt[\LARGE{n}]{\left( X_1 \times X_2 \times \ldots \times X_n \right)} = \exp(AM(\log(X))$$

Practically a GM difference is a multiplicative difference: you pay X% of a premium in interest when assuming a loan, your hemoglobin levels decrease X% after starting metformin, the failure rate of springs increase X% as a fraction of the width. In all of these instances, a raw mean difference makes less sense.
Log transforming estimates a geometric mean difference. If you log transform an outcome and model it in a linear regression using the following formula specification: log(y) ~ x, the coefficient $\beta_1$ is a mean difference of the log outcome comparing adjacent units of $X$. This is practically useless, so we exponentiate the parameter $e^{\beta_1}$ and interpret this value as a geometric mean difference. 
For instance, in a study of HIV viral load following 10 weeks administration of ART, we might estimate prepost geometric mean of $e^{\beta_1} = 0.40$. That means whatever the viral load was at baseline, it was on average 60% lower or had a 0.6 fold decrease at follow-up. If the load was 10,000 at baseline, my model would predict it to be 4,000 at follow-up, if it were 1,000 at baseline, my model would predict it to be 400 at follow-up (a smaller difference on the raw scale, but proportionally the same).
This is an important distinction from other answers:
The convention of multiplying the log-scale coefficient by 100 comes from the approximation $\log(x) \approx 1-x$ when $1-x$ is small. If the coefficient (on the log scale) is say 0.05, then $\exp(0.05) \approx 1.05$ and the interpretation is: a 5% "increase" in the outcome for a 1 unit "increase" in $X$. However, if the coefficient is 0.5 then $\exp(0.5) = 1.65$ and we interpret this as a 65% "increase" in $Y$  for a 1 unit "increase" in $X$. It is NOT a 50% increase.
Suppose we log transform a predictor: y ~ log(x, base=2). Here, I am interested in a multiplicative change in $x$ rather than a raw difference. I now am interested in comparing participants differing by 2 fold in $X$. Suppose for instance, I am interested in measuring infection (yes/no) following exposure to blood-borne pathogen at various concentrations using an additive risk model. The biologic model may suggest that risk increases proportionately for every doubling of concentration. Then, I do not transform my outcome, but the estimated $\beta_1$ coefficient is interpreted as a risk difference comparing groups exposed at two-fold concentration differences of infectious material.
Lastly, the log(y) ~ log(x) simply applies both definitions to obtain a multiplicative difference comparing groups differing multiplicatively in exposure levels.
