What is 4log_{b}(m)+log_{b}(n) ?

The solution to 4log_{b}(m)+log_{b}(n) is log_{b}(m^4n)

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simplify 4log_{b}(m)+log_{b}(n)

simplify

4 lo g_{b} (m) + lo g_{b} (n)

lo g_{b} (m^{4} n)

Solution steps

4 lo g_{b} (m) + lo g_{b} (n)

Apply log rule:

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

4 lo g_{b} (m) = lo g_{b} (m^{4})

= lo g_{b} (m^{4}) + lo g_{b} (n)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{b} (m^{4}) + lo g_{b} (n) = lo g_{b} (m^{4} n)

= lo g_{b} (m^{4} n)

ln (\frac{1}{2} \cdot x^{0.5} \cdot y^{0.5})

lo g_{5} (25 (x + y)^{5})

ln (\frac{x ^{2} + 1}{( x + 3 ) ( x - 2 )})

- \frac{1}{0.07} ln (\frac{1}{100})

What is 4log_{b}(m)+log_{b}(n) ?
The solution to 4log_{b}(m)+log_{b}(n) is log_{b}(m^4n)