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Just testing ${\log _a}n = \frac{{{{\log }_b}n}}{{{{\log }_b}a}}$. And one more time $${\log _a}n = \frac{{{{\log }_b}n}}{{{{\log }_b}a}}$$
Just testing ${\log _a}n = \frac{{{{\log }_b}n}}{{{{\log }_b}a}}$. And one more time $${\log _a}n = \frac{{{{\log }_b}n}}{{{{\log }_b}a}}$$