What is expand log_{10}(\sqrt[3]{3r^2s}) ?

The solution to expand log_{10}(\sqrt[3]{3r^2s}) is 1/3 log_{10}(3s)+2/3 log_{10}(r)

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expand log_{10}(\sqrt[3]{3r^2s})

expand

\frac{1}{3} lo g_{10} (3 s) + \frac{2}{3} lo g_{10} (r)

Solution steps

Rewrite as

= lo g_{10} ((3 r^{2} s)^{\frac{1}{3}})

Apply log rule

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

assuming

x \geq 0

= \frac{1}{3} lo g_{10} (3 r^{2} s)

Apply log rule:

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

lo g_{10} (3 r^{2} s) = lo g_{10} (3 s) + lo g_{10} (r^{2})

= \frac{1}{3} (lo g_{10} (r^{2}) + lo g_{10} (3 s))

Apply log rule

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

assuming

x \geq 0

= \frac{1}{3} (2 lo g_{10} (r) + lo g_{10} (3 s))

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{3}, b = lo g_{10} (3 s), c = 2 lo g_{10} (r)

= \frac{1}{3} lo g_{10} (3 s) + \frac{1}{3} \cdot 2 lo g_{10} (r)

= \frac{1}{3} lo g_{10} (3 s) + 2 \cdot \frac{1}{3} lo g_{10} (r)

2 \cdot \frac{1}{3} lo g_{10} (r) = \frac{2}{3} lo g_{10} (r)

= \frac{1}{3} lo g_{10} (3 s) + \frac{2}{3} lo g_{10} (r)

(x^{4} - 4 x + 7) d x

\frac{x + 6}{x ( x - 5 )}

(1 - i 3) (1 - i 3)

ln (x - 1) (x + 2)

(x - 2)^{6}

What is expand log_{10}(\sqrt[3]{3r^2s}) ?
The solution to expand log_{10}(\sqrt[3]{3r^2s}) is 1/3 log_{10}(3s)+2/3 log_{10}(r)