Let $\alpha$ be an ordinal and let $n\in \omega$ ($\omega$ is the least infinite ordinal). Suppose that $\alpha\geq\omega$ and $\gamma<\log\left(\alpha\right)$. I want to prove that $$\log\left(\alpha+\left(\omega^{\gamma}\cdot n\right)\right)=\log\left(\alpha\right).$$
For any ordinal $\beta$, I define $\log\left(\beta\right)$ as follows: $\beta$ can be written uniquely in the form $\beta=\omega^{\eta}\cdot m+\zeta$, $0<m<\omega$ and $\zeta<\omega^{\eta}$. Then $\log\left(\beta\right)$ is defined to be $\eta$.
I have already seen that $\log$ (considered as a function of ordinals) is increasing, which means that $$\log\left(\alpha+\left(\omega^{\gamma}\cdot n\right)\right)\geq\log\left(\alpha\right).$$
I just need to show that the case $$\log\left(\alpha+\left(\omega^{\gamma}\cdot n\right)\right)>\log\left(\alpha\right)$$ is impossible, but I'm not sure how to do that.
First thing - you can change your definition of $\log{\alpha}$ to be the least ordinal $\mu$ such $\alpha \leq \omega^{\mu}n$ for some sufficiently large $n$.
We have: $$\alpha = \omega^{\nu}n + \zeta$$ $$\beta = \omega^{\mu}m + \xi, \, \nu > \mu$$
We want to prove that $$\alpha + \beta \leq \omega^{\nu}q$$ For sufficiently large $q$.
But we can simply write: $$\alpha + \beta = \omega^{\nu}n + \zeta + \omega^{\mu} + \xi \leq \omega^{\nu}(n+1) + \omega^{\mu}(m+1) \leq \omega^{\nu}(n+1) + \omega^{\nu}(m+1) \leq \omega^{\nu}(n+m+2) $$
Then you just apply our equivalent definition of $\log$ to get upperbound for $\log{(\alpha + \beta)}$
Tbh your question is weirdly specific, because it can be stated simply as:
Let $\alpha$ > $\beta$ be ordinals, then $\log{(\alpha + \beta)} = \log{\alpha}$