Prove O(log(a+b))=O(log(a)+log(b))

To prove: $\mathbf{O}(\log{(a+b)})=\mathbf{O}(\log{a}+\log{b})$
We need to have:
$k_1\log{(a+b)} \le \log{a}+\log{b} \le k_2\log{(a+b)}$
is equivalent to:
$k_1\log{(a+b)} \le \log{(a*b)} \le k_2\log{(a+b)}$

Let e’s exponent of both side, we have:
$(a+b)^{k_1} \le a*b \le (a+b)^{k_2}$

Let $k_1=1, A=1, B = 1$, we know $\forall{a>A}, \forall{b>B}$ and $(a+b) \le a*b$ is always satisfied.

Let $k_2=2, A=1, B = 1$, we know $\forall{a>A}, \forall{b>B}$ $a*b \le (a+b)^2 \iff a*b \le a^2+2ab+b^2 \iff 0 \le a^2+ab+b^2$ is always satisfied.