What is 5log_{b}(m)-log_{b}(n) ?

The solution to 5log_{b}(m)-log_{b}(n) is log_{b}((m^5)/n)

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simplify 5log_{b}(m)-log_{b}(n)

simplify

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

lo g_{b} (\frac{m ^{5}}{n})

Solution steps

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

Apply log rule:

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

5 lo g_{b} (m) = lo g_{b} (m^{5})

= lo g_{b} (m^{5}) - 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^{5}) - lo g_{b} (n) = lo g_{b} (\frac{m ^{5}}{n})

= lo g_{b} (\frac{m ^{5}}{n})

3 ln (x) + 2 ln (y) - 6 ln (z)

lo g_{10} (x) - lo g_{10} (2)

lo g_{10} (100 y^{4} \times \frac{y + 5}{y - 5})

lo g_{4} (\frac{16}{x + 7})

\frac{3}{2} \cdot lo g_{3} (a) - \frac{1}{4} \cdot lo g_{3} (16 a^{2})

What is 5log_{b}(m)-log_{b}(n) ?
The solution to 5log_{b}(m)-log_{b}(n) is log_{b}((m^5)/n)