Your model and its estimates posit that
$$\sqrt{Y} = 2.1014 D - 3.0147 + \varepsilon$$
where $D$ is Dose.Back (or its logarithm) and $\varepsilon$ is a random variable of zero expectation whose standard deviation is approximately $16.28.$ Squaring both sides gives
$$Y = (2.1014 D - 3.0147 + \varepsilon)^2.$$
Adding $0.01$ to $D$ yields the value
$$(2.1014 (D + 0.01) - 3.0147 + \varepsilon')^2.$$
The difference is
$$2(2.1014 D - 3.0147 + \varepsilon)(\varepsilon' - \varepsilon + (0.01)(2.1014)) + (\varepsilon' - \varepsilon + (0.01)(2.1014))^2.$$
This expression, as well as its expectation, are complicated. Let us therefore focus on the simpler question of how the expectation of $Y$ varies with $D$. Note that
$$\eqalign{
\mathbb{E}(Y) &= \mathbb{E}\left(2.1014 D - 3.0147 + \varepsilon\right)^2 \\
&= (2.1014D - 3.0147)^2 + 2(2.1014D - 3.0147) \mathbb{E}(\varepsilon) + \mathbb{E}(\varepsilon^2) \\
&=(2.1014D - 3.0147)^2 + 0 + (16.28)^2.
}$$
(This result is of considerable interest in its own right because it reveals the role played by the mean squared error in interpreting the relationship between $D$ and $Y$.)
When $0.01$ is added to $D$ the value of $\mathbb{E}(Y)$ increases by
$$2(2.1014)(2.1014D - 3.0147)(0.01) + 2.1014(0.01)^2.$$
The last term $2.1014(0.01)^2 \approx 0.0002$ is so small compared to the squared errors (with their typical value of $16.28$) that we may neglect it. In this case, to a good approximation, this fitted model associates an (additive) increase in $D$ of $0.01$ with an increase in $Y$ of
$$2(2.1014)(2.1014D - 3.0147)(0.01) = 0.0883176 D - 0.126.$$
When $D$ is the natural logarithm of some quantity $d$, a 1% multiplicative increase in $d$ causes a value of approximately $0.01$ to be added to $D$, because
$$\log(1.01 d) = \log(1.01) + \log(d) = \left(0.01 - (0.01)^2/2 + \cdots\right) + D \approx 0.01 + D.$$
If you used a logarithm to another base $b$, entailing $D = \log_b(d) = \log(d)/\log(b),$ then a 1% multiplicative increase in $d$ causes a value of approximately $(0.01)/\log(b)$ to be added to $D$, so everywhere "$0.01$" occurs in the preceding formulas you must use $(0.01/\log(b))$ instead.
The interpretation of coefficients associated with log-transformed independent variables is straightforward. You now have log units, which depend on the choice of basis for the logarithm. For natural log as in your example, an $e$-fold change in (Independent_var_1 + 1) is associated with the indicated change in the dependent variable. It might be simpler for a reader to understand if you to use base-2 or base-10 logarithms, so that the regression coefficient represents a doubling or a 10-fold increase in (Independent_var_1 + 1).
You might want to consider the suggestion provided here for an alternate way of dealing with the 0-value problem in logarithmic transformations, which handles cases having 0 values of an independent variable separately from cases having positive values.
Best Answer
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").