What is 3log_{b}(m)-2log_{b}(n) ?

The solution to 3log_{b}(m)-2log_{b}(n) is log_{b}((m^3)/(n^2))

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simplify 3log_{b}(m)-2log_{b}(n)

simplify

3 lo g_{b} (m) - 2 lo g_{b} (n)

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

Solution steps

3 lo g_{b} (m) - 2 lo g_{b} (n)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 lo g_{b} (m) = lo g_{b} (m^{3})

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{b} (n) = lo g_{b} (n^{2})

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

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

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

= lo g_{b} (\frac{m ^{3}}{n ^{2}})

lo g_{10} (x - 1) + lo g_{10} (x + 1)

lo g_{10} (\frac{x ^{5}}{y ^{2}})

ln (e (x + 5))

ln (\frac{e ^{4}}{9})

lo g_{2} (x) - lo g_{2} (w) + lo g_{2} (z)

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