What is 1/5 [log_{4}(x)+log_{4}(3)]-[log_{4}(y)] ?

The solution to 1/5 [log_{4}(x)+log_{4}(3)]-[log_{4}(y)] is log_{4}((\sqrt[5]{x}\sqrt[5]{3})/y)

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simplify 1/5 [log_{4}(x)+log_{4}(3)]-[log_{4}(y)]

simplify

\frac{1}{5} [lo g_{4} (x) + lo g_{4} (3)] - [lo g_{4} (y)]

Solution steps

\frac{1}{5} (lo g_{4} (x) + lo g_{4} (3)) - (lo g_{4} (y))

Apply rule:

- (a) = - a

- (lo g_{4} (y)) = - lo g_{4} (y)

= \frac{1}{5} (lo g_{4} (x) + lo g_{4} (3)) - lo g_{4} (y)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{4} (x) + lo g_{4} (3) = lo g_{4} (x \cdot 3)

= \frac{1}{5} lo g_{4} (x \cdot 3) - lo g_{4} (y)

Apply log rule:

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

\frac{1}{5} lo g_{4} (x \cdot 3) = lo g_{4} ((x \cdot 3)^{\frac{1}{5}})

= lo g_{4} ((x \cdot 3)^{\frac{1}{5}}) - lo g_{4} (y)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{4} ((x \cdot 3)^{\frac{1}{5}}) - lo g_{4} (y) = lo g_{4} (\frac{( x \cdot 3 ) ^{\frac{1}{5}}}{y})

= lo g_{4} (\frac{( x \cdot 3 ) ^{\frac{1}{5}}}{y})

Simplify

\frac{( x \cdot 3 ) ^{\frac{1}{5}}}{y} : \frac{x ^{\frac{1}{5}} \cdot 3 ^{\frac{1}{5}}}{y}

= lo g_{4} (\frac{x ^{\frac{1}{5}} \cdot 3 ^{\frac{1}{5}}}{y})

Apply exponent rule:

ln (x) + ln (x - 2)

lo g_{b} (\frac{x y}{z})

lo g_{4} (\frac{3}{x})

lo g_{10} (3 x^{3})

lo g_{12} (9) + lo g_{12} (14)

What is 1/5 [log_{4}(x)+log_{4}(3)]-[log_{4}(y)] ?
The solution to 1/5 [log_{4}(x)+log_{4}(3)]-[log_{4}(y)] is log_{4}((\sqrt[5]{x}\sqrt[5]{3})/y)