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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

ln (\frac{2 x}{y})

ln (∣ 11.9 - 1 ∣^{2} ∣ 2 \cdot 11.9 + 1 ∣^{\frac{1}{2}})

7 lo g_{10} (y) + 2 lo g_{10} (5 y + 3)

0.7 ln (\frac{14 m}{14 x + 3 y})

- lo g_{10} (1.9 \cdot 1 0^{- 9})

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