What is log_{4}(1/16 x^2y^3) ?

The solution to log_{4}(1/16 x^2y^3) is -2+log_{2}(x)+3log_{4}(y)

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simplify log_{4}(1/16 x^2y^3)

simplify

lo g_{4} (\frac{1}{16} x^{2} y^{3})

- 2 + lo g_{2} (x) + 3 lo g_{4} (y)

Solution steps

lo g_{4} (\frac{1}{16} x^{2} y^{3})

\frac{1}{16} = 4^{- 2}

= lo g_{4} (4^{- 2} x^{2} y^{3})

Apply log rule:

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

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

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

Apply log rule:

lo g_{a} (a^{b}) = b

lo g_{4} (4^{- 2}) = - 2

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

Simplify

lo g_{4} (x^{2} y^{3}) : lo g_{2} (x) + 3 lo g_{4} (y)

= - 2 + (lo g_{2} (x) + 3 lo g_{4} (y))

Apply rule:

a + (b + c) = a + b + c

- 2 + (lo g_{2} (x) + 3 lo g_{4} (y)) = - 2 + lo g_{2} (x) + 3 lo g_{4} (y)

= - 2 + lo g_{2} (x) + 3 lo g_{4} (y)

ln (\frac{5}{3}) + 4

lo g_{7} (9) + lo g_{7} (8)

f (x) = lo g_{10} (2 x) - lo g_{10} (8 x^{3})

ln (x y z^{6})

\frac{ln ( 2 )}{ln ( 2 ) - 2 ln ( 3 )}

What is log_{4}(1/16 x^2y^3) ?
The solution to log_{4}(1/16 x^2y^3) is -2+log_{2}(x)+3log_{4}(y)