Converting log standard deviation in terms of the original units We run regression specifications of the form,
lm(outcome_log_s ~ x1_s + x1_x^2)
In which the outcome variable is logged and standardized. We want to be able to convey what, for example, a coefficient of ".1" on one of our covariates means is in terms of the original y-scale.  How would we go about doing this given standard deviation information from mean centered and log mean centered versions of the same variable
 A: TL;DR
Additive changes in log imply multiplicative effect. The scaling of the output also plays a role as it changes the scale of the multiplicative effect. That is, if change of 1 in the units of $X$ corresponds to a change of 0.69 in the units of scaled log $Y$, then this implies $e^{0.69 \cdot \text{sd}(\log(Y))}$.
As standard deviation is not affected by centering, we can use sd(data$outcome_log_m) as the value of $\text{sd}(\log(Y))$. Thus, we get a multiplicative affect of $ e^{0.69 \cdot 0.8339012 } \approx 1.70$. 
In terms of interpertation, this means 70% increase in the value measured.
Explanation
Assume we transform the data as $$ Z_{i} = \frac{\log(Y_{i}) - \overline{\log(Y)}}{\text{sd}(\log(Y))}$$
and model it as 
$$ Z_i = \beta^{\top} X_i + \eta_{i}$$
where $\eta_{i}$ models the noise.
Assume that for a specific point $i$, $X_{i}$ changes to $X_{i,\varepsilon} = X_{i} + \varepsilon$, without affecting the value of $\eta$. This implies $$Z_{i,\varepsilon} - Z_{i} = \beta^{\top} \varepsilon.$$ On the other hand, assuming the mean and standard deviation are close to the true mean and standard deviation (so the $\varepsilon$ won't change them a lot), $$Z_{i,\varepsilon} - Z_{i} = \frac{\log(Y_{i,\varepsilon}) - \overline{\log(Y)}}{\text{sd}(\log(Y))} - \frac{\log(Y_{i}) - \overline{\log(Y)}}{\text{sd}(\log(Y))} = \frac{1}{\text{sd}(\log(Y))} \log\left(\frac{Y_{i,\varepsilon}}{Y_{i}}\right).$$
Together, they imply,
$$ \log\left(\frac{Y_{i,\varepsilon}}{Y_{i}}\right) = \beta^{\top}\varepsilon \cdot \text{sd}\left(\log(Y)\right)$$
which can be written as
$$ Y_{i,\varepsilon} = e^{\beta^{\top}\varepsilon \cdot \text{sd}\left(\log(Y)\right)} Y_{i}$$
Precise calculation would take into account the different sources of error - from the mean calculation, standard deviation calculation and estimation error to create a confedence interval.
