<s>[INST] Solving $2^n = 10^6$ for $n$ I am trying to solve $2^n = 10^6$ for $n$. Is there some way that I can solve this analytically (say, using base $2$ logarithms), or must I solve it numerically? Thank you.
 [/INST] Well, sure.  $\log_2 2^n = \log_2 10^6$
$n = 6\log_2 10$ or 
$\log_{10} 2^n = \log_{10} 10^6$ 
$n \log_{10} = 6$
$n = \frac {6}{\log_{10}2}$ or 
$\log_b 2^n = \log_b 10^6$
$n\log_b 2^n = 6\log_b 10$
$n = 6\frac {\log_b 10}{\log_b 2}$ for any $b > 0; b\ne 1$ are all acceptable answers.
But it's worth noting that 
1)  $\frac {\log_b 10}{\log_b 2} = \log_2 {10}$ always.  that's called the change of base formula and a special case is $\frac 1{\log_{10}  2}=\frac {\log_{10} 10}{\log_{10} 2} = \log_2 {10}$.
2)  $\log_2 10= \log_2 (2*5) = \log_2 2 + \log_2 5 = 1 + \log_2 5$.
So $2^n = 10^6 = 2^6*5^6$
$\log_2 2^n = \log_2 (2^6*5^6)$
$n = \log_2 2^6 + \log_2 5^6 = 6 + 6\log_2 5=6(1 + \log_2 5)$.
All acceptable answers.
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