2
You solved your own question. Quite literally the bounds you gave (with $c=1$ or $c=2$) should suffice
1
Roughly speaking:
$\begin{align*}
\log_2 (2^n)
&= n \\
\log_2(10^n + n)
&> \log_2(10^n) \\
&= n \log_2 10
\end{align*}$
As $\log_2 10 = 3.3219$, the later is larger.
For rough comparisons, you can discard "lower terms" with impunity. In any case, often "asymptotic growth rate" applies only for very large $...
1
Compute $\lim_{n\to\infty}\frac{\log(n+10^n)}{2^{\log(n)}}$. This is
$$\begin{align}\lim_{n\to\infty}\frac{\log(n+10^n)}{2^{\log(n)}}&=\lim_{n\to\infty}\frac{\log(n+1+10^{n+1})-\log(n+10^n)}{1}\\
&=\lim_{n\to\infty}\log\left(\frac{(n+1)+10^{n+1}}{n+10^n}\right)\\
&=\lim_{n\to\infty}\log\left(\frac{(n+1)/10^n+10}{n/10^n+1}\right)\\
&=\log(10)\\...
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