Definition: What Is The "Trend Growth Rate"? I am a bit unsure about how the "trend growth rate" is defined in macroeconomics. Is it to correct to say that if a time series $\{y_t\}$ may be decomposed as $$y_t=d_t+c_t,$$ where $d_t$ is a deterministic trend and $c_t$ a cyclical component, then $$(d_{t+1}-d_t)/d_t$$ is the "trend growth rate", i.e. the growth rate in $y_t$ assuming $c_t=0$? And if it can be decomposed as $$\log(y_t)=d'_t+c'_t,$$ with another trend and cyclical component, then $$[\exp(d'_{t+1})-\exp(d'_t)]/\exp(d'_t)$$ is the "trend growth rate", i.e. the growth rate in $y_t$ assuming $c'_t=0$?

Sometimes I feel that the definition is $(d_{t+1}-d_t)/d_t$ and $(d'_{t+1}-d'_t)/d'_t$ in the two cases, referring to the growth rate in the trend component and not to the growth rate in the dependent variable $y_t$ ignoring the cyclical component (i.e., setting $c_t=0$ for all $t$). 

That is, I interpret "trend growth rate" as the growth rate in $y_t$ ignoring the cyclical component. Is that a correct interpretation?
Any time series can be decomposed in four terms:


trend: the underlying change in the variable if no other factor is present. In the simplest case, think of an upward slopping line. The growth rate of this trend is it's slope. 
cycle: the cyclical component of the series, defined by ups and downs like a wave, with a certain "frequency" and "amplitude". Think of this like a wave.
seasonal component: usually a within year cycle. The standard textbook example is that of ice cream sales, which pick up in summer. This is different from the longer cycle described above, and perhaps less smooth (e.g. as ice cream sales are low 3/4 of the year)
noise: natural randomness due to measurement errors or other unobserved yet minor factors, including shocks (e.g. a very hot weather in a week, increasing ice cream sales)


For a series $y_t$, this is something like

$$ y_t = d_t + c_t + s_t + e_t $$

Importantly, all the variables are defined in the same units by construction. E.g. if this is GDP, they are all in value added units (real, nominal, whatever). 

A general case of a decomposition is (assume just trend and cycle):

$$ f(y_t) = d^{'}_{t} + c^{'}_{t} $$

In this case, the trend growth of $f(y_t)$ is the growth rate of $d^{'}_t$. But this is not the same as the trend growth rate of $y_t$. 

Consider an example. Let us construct a series $y_t$, where $d_t=d_0(1+g)^t$ and $c_t=sin(t)$, and $y_t=d_t+c_t$. This is a trend-cycle decomposition where the growth rate of the trend is $g$:

$$ \frac{d_{t+1}-d_t}{d_t} = \frac{d_0(1+g)^{t+1}-d_0(1+g)^t}{d_0(1+g)^t} = \cdots = g $$

Now, transform the process into log, and let us postulate a new decomposition:

$$ \log y_t = D_t + C_t $$

where $D_t = D_0(1+G)^t$, and $C_t$ is an unknown function. This is by definition correct, but $C_t$ might be an irregular, perhaps not algebraically defined function (like $sin(t)$). In this new case, it is correct, by definition, that the trend growth rate of the log of $y_t$ is $G$, computed with the standard formula. 

The central question is: can we find a general method to calculate the trend growth rate of $y_t$ ($g$) using only some transformation of $D_t$? (as in your attempt using the exponential). In other words, can we use only $D_0(1+G)^t$ to obtain $g$?

(Here I highlight the point of why would you ever go down this route, if you can decompose $y_t$ in levels and solve your problem immediately; but anyway, let us continue).

The answer seems to be negative. It is trivial to see that

$$ e^{D_0(1+G)t} e^{C_t} = d_0(1+g)^t + sin(t) $$

where you cannot find a direct function between $G$ and $g$ which does not involve the use of $C_t$ and $sin(t)$. In other words, you need to use the information of both decompositions to solve the problem. But this is circular. To make the link between logs and levels, you need to make the decomposition in levels. But if you do the latter, the answer is there already. No need to go further.  

Finally, consider the very special case where the series $y_t$ only has a constant growth rate (trend):

$$ y_t = d_t = d_0(1+g)^t $$

Then,

$$ \log y_t = D_t = D_0(1+G)^t = \log \left(d_0(1+g)^t\right) $$

The following formula recovers the original growth rate of $y_t$ in terms of the trend from the series in log:

$$ \dfrac{e^{D_{t+1}} - e^{D_t}}{e^{D_t}} = \frac{d_{t+1} - d_t}{d_t} = \cdots = g $$

The result can be generalised to any $f(y_t)$, where the transformation of the growth rate calculation has to be $f^{-1}(\cdot)$.