What is 3log_{4}(3)-log_{4}(12+2) ?

The solution to 3log_{4}(3)-log_{4}(12+2) is log_{4}(27/14)

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simplify 3log_{4}(3)-log_{4}(12+2)

simplify

3 lo g_{4} (3) - lo g_{4} (12 + 2)

lo g_{4} (\frac{27}{14})

Decimal Notation

0.47376 \dots

Solution steps

3 lo g_{4} (3) - lo g_{4} (12 + 2)

Apply log rule:

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

3 lo g_{4} (3) = lo g_{4} (3^{3})

= lo g_{4} (3^{3}) - lo g_{4} (12 + 2)

Apply log rule:

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

lo g_{4} (3^{3}) - lo g_{4} (12 + 2) = lo g_{4} (\frac{3 ^{3}}{12 + 2})

= lo g_{4} (\frac{3 ^{3}}{12 + 2})

\frac{3 ^{3}}{12 + 2} = \frac{27}{14}

= lo g_{4} (\frac{27}{14})

\frac{sin ( 2 x )}{sin ( 3 x )}

64 m^{9} + 64 n^{16}

- \frac{1}{cos ( x )}

\frac{x ^{- 3} y}{x y ^{2}}

\frac{( \frac{1}{5} ) ^{3}}{5 ^{- 2}}

What is 3log_{4}(3)-log_{4}(12+2) ?
The solution to 3log_{4}(3)-log_{4}(12+2) is log_{4}(27/14)