${}^a\log b=c\to a^c=b$

$a>0,b>0$ dan $a\ne 1$

${}^a\log c^p=p.{}^a\log c$

${}^{a^q}\log c=\frac{1}{q}.{}^a\log c$

${}^{a^q}\log c^p=\frac{p}{q}.{}^a\log c$

${}^c\log a+{}^c\log b={}^c\log a.b$

${}^c\log a-{}^c\log b={}^c\log {\displaystyle \frac{a}{b}}$

${}^a\log b={\displaystyle \frac{1}{{}^b\log a}}$

${}^a\log b.{}^b\log c={}^a\log c$

$(c{)}^{{}^c\log a}=a$