What is log_{2}(n/(3^i)) ?

The solution to log_{2}(n/(3^i)) is log_{2}(n)-ilog_{2}(3)

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simplify log_{2}(n/(3^i))

simplify

lo g_{2} (\frac{n}{3 ^{i}})

lo g_{2} (n) - i lo g_{2} (3)

Solution steps

lo g_{2} (\frac{n}{3 ^{i}})

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

lo g_{2} (\frac{n}{3 ^{i}}) = lo g_{2} (n) - lo g_{2} (3^{i})

= lo g_{2} (n) - lo g_{2} (3^{i})

Simplify

lo g_{2} (3^{i}) : i lo g_{2} (3)

= lo g_{2} (n) - i lo g_{2} (3)

ln (2) + ln (\frac{4}{3})

\frac{2}{3} lo g_{4} (x) + lo g_{4} (y)

ln (x) - ln (10)

lo g_{3} (\frac{w}{9})

What is log_{2}(n/(3^i)) ?
The solution to log_{2}(n/(3^i)) is log_{2}(n)-ilog_{2}(3)