What is expand log_{3}(\sqrt[5]{mn^2}) ?

The solution to expand log_{3}(\sqrt[5]{mn^2}) is 1/5 log_{3}(m)+2/5 log_{3}(n)

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expand log_{3}(\sqrt[5]{mn^2})

expand

\frac{1}{5} lo g_{3} (m) + \frac{2}{5} lo g_{3} (n)

Solution steps

Rewrite as

= lo g_{3} ((m n^{2})^{\frac{1}{5}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{5} lo g_{3} (m n^{2})

Simplify

lo g_{3} (m n^{2}) : lo g_{3} (m) + 2 lo g_{3} (n)

= \frac{1}{5} (lo g_{3} (m) + 2 lo g_{3} (n))

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{5}, b = lo g_{3} (m), c = 2 lo g_{3} (n)

= \frac{1}{5} lo g_{3} (m) + \frac{1}{5} \cdot 2 lo g_{3} (n)

= \frac{1}{5} lo g_{3} (m) + 2 \cdot \frac{1}{5} lo g_{3} (n)

2 \cdot \frac{1}{5} lo g_{3} (n) = \frac{2}{5} lo g_{3} (n)

= \frac{1}{5} lo g_{3} (m) + \frac{2}{5} lo g_{3} (n)

(- 3 e^{- t}) (- 5 e^{4 t} + e^{- 2 t})

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

2^{3}

(a + b - ab) (c + d - c d) (e + f - e f)

2^{8}

What is expand log_{3}(\sqrt[5]{mn^2}) ?
The solution to expand log_{3}(\sqrt[5]{mn^2}) is 1/5 log_{3}(m)+2/5 log_{3}(n)