To explore whether a model is linear in a particular element, such as $\beta_1$ (for example), let's adopt a notation that focuses on this element. Write
$$f(\theta) = \log(\theta X_{i1}) + \beta_2 X_{i2} + \epsilon_i,\tag{1}$$
thereby suppressing all other variables and parameters in the definition of $f$ and generically referring to the parameter $\beta_1$ as "$\theta.$" This ought to make it obvious how the following approach applies to any parameter for any model.
Let's be clear about the objective.
Linearizing $f$ means there is a one-to-one transformation $u,$ which converts a new parameter $\gamma$ into the original parameter $\theta=u(\gamma),$ for which $g(\gamma)=f(u(\gamma))$ is a linear function of $\gamma$ wherever it is defined.
A fairly general meaning of "$g$ is linear" is that $g$ is differentiable on an open set containing its domain and has a constant derivative $g^\prime.$ (This generality really is needed, because many statistical models constrain some of their parameters. For instance, the parameter $\sigma$ in the familiar Normal$(\mu,\sigma^2)$ family of distributions is usually constrained to $\sigma\gt 0.$)
This approach allows us to use the machinery of Calculus to decide the question whether $f$ can be linearized. To apply it, we usually assume the reparameterization is itself differentiable. The Chain Rule implies $g$ is differentiable. Let $C$ name the (as yet unknown) constant derivative of $g.$ The Chain Rule also supplies a formula for the derivative,
$$C=g^\prime(\gamma) = f^\prime(u(\gamma))\, u^\prime(\gamma).\tag{2}$$
(Bear in mind that the model is undefined if $X_{i1}=0,$ so we may assume these numbers are nonzero.)
The original formula $(1)$ provides the information needed to differentiate $f,$
$$f^\prime(\theta) = \frac{X_{i1}}{\theta X_{i1}} = \frac{1}{\theta}.$$
Combine this with $(2)$ to obtain an equation for the unknown reparameterization $u:$
$$u^\prime(\gamma) = \frac{C}{f^\prime(u(\gamma))} = C u(\gamma),\tag{3}$$
whence
$$\frac{du}{u}\left(\gamma\right) = Cd\gamma$$
with general solution
$$\log\, \lvert \theta \rvert = \log\, \lvert u(\gamma) \rvert = C \gamma + C_0,$$
equivalent to
$$\theta = \pm e^{C\gamma + C_0}$$
for some additional constant $C_0.$ The sign to choose depends on whether all the $X_{i1}$ are positive or all are negative. (The model is undefined if the signs of these variables vary.)
To be quite explicit, in the case all the $X_{i1}$ are positive, we may choose $C=1$ and $C_0=0$ so that the model is
$$Y_i = \gamma \log(X_{i1}) + \beta_2 X_{i2} + \epsilon_i;\quad \beta_1 = e^\gamma$$
and when all the $X_{i1}$ are negative, the model can be written
$$Y_i = \gamma \log(-X_{i1}) + \beta_2 X_{i2} + \epsilon_i;\quad \beta_1=-e^\gamma.$$
However, being able to find an explicit solution to $(3)$ is not important for us: the mere demonstration that there exists some solution suffices to show that $f$ is linearizable. Note, too, that you didn't need to know much about logarithms to arrive at $(3),$ nor was it necessary to be clever or have some kind of insight. The work was purely mechanical.
For more on the many different meanings of "linear" model, please see my post at https://stats.stackexchange.com/a/148713/919.
