# Which is more efficient? lg(n+10^n) higher than 2^lgn [duplicate]

Based on the order by asymptotic growth rate which is more efficient?

• It could depend on what base log you are using. Could you be more specific in the question? – Richie Yeung Jul 20 '20 at 14:05
• it is base log 2 – Veree Jul 20 '20 at 14:08
• If it is base 2 then the second term might as well be n. – Richie Yeung Jul 20 '20 at 14:12

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)\\ &>1 \end{align}

where the first equation was using Stolz-Cesaro theorem (L'Hospital for sequences).

That this limit is larger than $$1$$ implies that for $$n$$ large enough $$\log(n+10^n)$$ becomes larger than $$2^{\log(n)}$$.

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 $$n$$, and is usually meant to hide constant factors (check e.g. Hildebrand's "Short Course on Asymptotics" for details on the notation and manipulation techniques). If so, both are $$\Theta(n)$$, and more details are needed to compare fairly.

• The last but one equality statement is false – droptop Jul 20 '20 at 17:24