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C23panweiming's blog
2024.1.21
C23panweiming LV 7 @ 2024-1-22 18:27:23

$a^na^m=a^{n+m}$
$(a^n)^m=a^{nm}$
$log_a n=x <=> a^x=n$
$log_a (nm)=log_a m*log_a n$
$log_a (n/m)=log_a n/log_a m$
$log_a m^n=n log_a m$
$log_a b=log_c a/log_c b$