Interpreting Standard Deviation of Natural Log Transformed Data I am interested in interpreting (back transforming) the effect of a one standard deviation (sd) increase in a log transformed on the non-transformed variable.
So let's say I have a variable Y:
Y= # of likes

ln(Y)= log transformed # of likes

mean(ln(Y))=7.7

sd(ln(Y))=0.8

Now I want to relate the  sd(ln(Y)) back to non-transformed units Y. In other words, how many (#) likes is a one standard deviation increase of the log transformed units( sd(ln(Y))=0.8) equal to?
I thought maybe I could simply compare the change in the transformed data mean with the standard units) and back transform using the exponential function
e^(7.7+0.8)- e^(7.7)≈2706

So a one standard deviation increase of the log-transformed variable translates to 2,706 likes. Is this ok? Or should I be using another formula to calculate this?
 A: The proposed interpretation in your last paragraph is incorrect -- that increase only applies at the mean. If you started lower, it would be a smaller increase and if you started higher it would be a larger increase.
$e^{a+0.8}- e^a=e^{a}(e^{0.8}- 1)\approx 1.2255 e^a$
It's better to think in terms of percentage increase.
$\frac{e^{a+0.8}- e^a}{e^a}\approx 1.2255$, or about 122.5% increase.
However, I am concerned about your use of logs on a count that could be zero (count of "likes").
A: 
So a one standard deviation increase of the log-transformed variable translates to 2,706 likes. Is this ok? 

You were careful to formulate your statement with 'increase of  of log-transformed variable' qualifier. I think this eliminates misunderstanding that could have occurred to a reader who may assume that you're trying to calculate the standard deviation of $Y$. You're clearly not trying to do that. You use a word 'translates', which is not a standard term thus indicating that you're not transforming variables and converting the statistics between these variables by 'standard' means.
Compare your procedure to what's described in "SAS/ETS 12.1 Users Guide", p.252 

The log transformation is often used to convert time series that are
  nonstationary with respect to the innovation variance into stationary
  time series. The usual approach is to take the log of the series in a
  DATA step and then apply PROC ARIMA to the transformed data. A DATA
  step is then used to transform the forecasts of the logs back to the
  original units of measurement. The confidence limits are also
  transformed by using the exponential function.

The highlighted [by me] sentence essentially describes what you're doing.
Hence, what you are doing is not wrong, whether it's right is an interesting question. It depends on the interpretations and the intended use.
One more thing (c) The estimator of the mean of the original variable $Y$ is not necessarily $e^{\overline{\ln Y}}$. I'm using a soft language here, because there's this seemingly obvious estimator
$$\hat\mu_Y=\exp\left(\hat\mu_{\ln Y}+\hat\sigma^2_{\ln Y}/2\right)$$
It is based on the exact relationship for log-normal distribution:
$$E[Y]=\exp\left(E[\ln Y]+\sigma^2_{\ln Y}/2\right)$$
However, this estimator is not always the best one in practice for the variance $\sigma^2_{\ln Y}$ is unknown, and has to be estimated. Once you start using the estimator of the variance, things get complicated, as shown in the empirical paper by Helmut Lutkepohl and Fang Xu. "The role of the log transformation in forecasting economic variables." Empirical
Economics, 42(3):619{638, 2012. 
The following, naive, estimator of the mean may end up being the best in such cases:
$$\hat\mu_Y'=\exp\left(\hat\mu_{\ln Y}\right)$$
I went to write about the means because when you talk about the 'translation' of the standard deviation increase, you need to mention what is the base. You assumed rather implicitly that the increase is from the point of the naive estimator above. As I wrote it is not wrong, but you have to clearly state that it's what you used, otherwise your reader may assume that you're correcting for the variance or that the 2,706 likes increase is from any point (which is not true). For instance, if you apply your equation to the base of 0, you get $$e^{0+0.8}-e^0=2.2$$ 
A: If I understand, you want the standard deviation of Y. The standard deviation of Y is NOT easily calculated from mean(ln(Y)) and sd(ln(Y)), so your formula is not okay. The easy solution is to ignore the log-transform when calculating the standard deviation of Y: i.e. sd(Y) or sd(e^ln(Y)). 
