What is expand log_{c}(\sqrt[8]{(x^2)/(y^3c^8)}) ?

The solution to expand log_{c}(\sqrt[8]{(x^2)/(y^3c^8)}) is 1/4 log_{c}(x)-3/8 log_{c}(y)-1

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expand log_{c}(\sqrt[8]{(x^2)/(y^3c^8)})

expand

\frac{1}{4} lo g_{c} (x) - \frac{3}{8} lo g_{c} (y) - 1

Solution steps

Rewrite as

= lo g_{c} ((\frac{x ^{2}}{y ^{3} c ^{8}})^{\frac{1}{8}})

Apply log rule

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

assuming

x \geq 0

= \frac{1}{8} lo g_{c} (\frac{x ^{2}}{c ^{8} y ^{3}})

Simplify

lo g_{c} (\frac{x ^{2}}{y ^{3} c ^{8}}) : 2 lo g_{c} (x) - lo g_{c} (c^{8} y^{3})

= \frac{1}{8} (2 lo g_{c} (x) - lo g_{c} (c^{8} y^{3}))

lo g_{c} (y^{3} c^{8}) = 3 lo g_{c} (y) + 8

= \frac{1}{8} (2 lo g_{c} (x) - (3 lo g_{c} (y) + 8))

- (3 lo g_{c} (y) + 8) : - 3 lo g_{c} (y) - 8

= \frac{1}{8} (2 lo g_{c} (x) - 3 lo g_{c} (y) - 8)

Distribute parentheses

= \frac{1}{8} \cdot 2 lo g_{c} (x) + \frac{1}{8} (- 3 lo g_{c} (y)) + \frac{1}{8} (- 8)

Apply minus-plus rules

+ (- a) = - a

= 2 \cdot \frac{1}{8} lo g_{c} (x) - 3 \cdot \frac{1}{8} lo g_{c} (y) - 8 \cdot \frac{1}{8}

Simplify

2 \cdot \frac{1}{8} lo g_{c} (x) - 3 \cdot \frac{1}{8} lo g_{c} (y) - 8 \cdot \frac{1}{8} : \frac{1}{4} lo g_{c} (x) - \frac{3}{8} lo g_{c} (y) - 1

= \frac{1}{4} lo g_{c} (x) - \frac{3}{8} lo g_{c} (y) - 1

\frac{2 ( 2 e ^{x} + cos ( x ) )}{x}

\frac{( L w ^{2} C - 1 ) ( L w ^{2} r _{2} C - r _{1} )}{w C ^{2}}

\frac{1}{( 1 + 3 x ) ^{3}}

\frac{( x - 2 ) ( x + 3 )}{( x + 5 ) ( x - 5 )}

(l a cos (a) + l b cos (b))^{2}

What is expand log_{c}(\sqrt[8]{(x^2)/(y^3c^8)}) ?
The solution to expand log_{c}(\sqrt[8]{(x^2)/(y^3c^8)}) is 1/4 log_{c}(x)-3/8 log_{c}(y)-1