Given the relationship: $$time = a • diameter^b for \ some \ constants \ a \ and \ b$$ How can we transform into a relationship which can be modelled by a simple linear regression, what assumptions are you making about errors in the original relationship?
My thoughts are that the given relationship is of the form y=ax^b, an exponential relationship. Hence, I would be able to take logs of both sides in order to get to a linear relationship. I.e.
$$log(time)=log(a)+b•log(diameter)$$
However, I am unsure if I am correct and cannot find any information on assumptions of errors in the original relationship for the transformation to occur.
What you started to do is correct $$\text{time}=a \, \text{diameter}^b\implies \log(\text{time})=\log(a)+b \log(\text{diameter})$$ but this is not a linear model (a model is said to be linear when it is linear with respect to all its parameters). It becommes linear if you let $c=\log(a)$.
But, in any manner, this is only the preliminary step because what has been measured is $\text{time}$ and not any of its possible transforms. The linear regression provides good (or at least reasonable) estimates of $a=e^c$ and $b$. From here, a nonlinear regression will work like a charm.
Up to this point, do not waste time asking yourself about assumptions of errors.
In fact, it is possible to show that staying with the linear regression is extremely close to the nonlinear regression for the minimization of the squares of relative errors.
When you use the linear model, the residue at point $i$ is $$r_i=\log\big[y_i^{\text{calc}}\big]-\log\big[y_i^{\text{exp}}\big]=\log\Bigg[\frac{y_i^{\text{calc}} } {y_i^{\text{exp}} }\Bigg]=\log\Bigg[1+\frac{y_i^{\text{calc}} -y_i^{\text{exp}}} {y_i^{\text{exp}} }\Bigg]$$
Now, if $$\frac{y_i^{\text{calc}} -y_i^{\text{exp}}} {y_i^{\text{exp}} } \ll 1$$ then, by Taylor $$\log\Bigg[1+\frac{y_i^{\text{calc}} -y_i^{\text{exp}}} {y_i^{\text{exp}} }\Bigg] \sim \frac{y_i^{\text{calc}} -y_i^{\text{exp}}} {y_i^{\text{exp}} } $$
Q.E.D.