Interpret Logarithmic Values like Mean and Stdlet's say i have a series of data containing...

Let $x_i$ your observations. Then the (observed) arithmetic mean ${\hat{\mu }}_{\log }$ of the logarithm of those is
$\frac{\sum _{i=1}^n\log (x_i)}{n}=\frac{\log \left(\prod _{i=1}^n{x}_i\right)}{n}=\log \left(\prod _{i=1}^n{x}_i^{1/n}\right)$
which is the logarithm of the (observed) geometric mean of the original values (i.e. $\exp ({\hat{\mu }}_{\log })$ is the geometric mean of the $x_i$).
The (observed) standard deviation of these values is
${\hat{\sigma }}_{\log }=\sqrt{\frac{\sum _{i=1}^n(\log (x_i)-{\hat{\mu }}_{\log }{)}^2}{n-1}}$
but the ${}^2$ makes it a bit more difficult to transform the way we transformed ${\hat{\mu }}_{\log }$ above. However, $\exp ({\hat{\sigma }}_{\log })$ does have an interpretation as a measure of deviation from the (geometric) mean, not in terms of difference but in terms of ratio.
When you have a lot of observations where exactly half of them are −1 and exactly half of them are 1, then the arithmetic mean is 0, and the standard deviation is a little larger than 1 (it tends to 1 as the number of observations grows), which is the observed deviation from the arithmetic mean in all cases.
In exactly the same way, for a long list of observations where exactly half are $\frac{1}{2}$ and exactly half are 2, the geometric mean $\exp ({\hat{\mu }}_{\log })$ is 1, and the "geometric standard deviation", $\exp ({\hat{\sigma }}_{\log })$ is a little larger than 2 (and tends to 2 as the number of observations grows), since all the observations deviate from the geometric mean by a factor of 2.
In your case, this means that you have observed a geometric mean of $\exp (0.55)\approx 1.73$, and a standard multiplicative deviation from that mean of $\exp (0.69)\approx 1.99$. So what would usually be an interval of "mean plus or minus a standard deviation" now becomes
$[1.73/1.99,1.73\cdot 1.99]=[0.87,3.46]$