You solved your own question. Quite literally the bounds you gave (with $c=1$ or $c=2$) should suffice


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 $...


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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