What is log_{4}(1/16 t^3v) ?

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

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simplify log_{4}(1/16 t^3v)

simplify

lo g_{4} (\frac{1}{16} t^{3} v)

- 2 + 3 lo g_{4} (t) + lo g_{4} (v)

Solution steps

lo g_{4} (\frac{1}{16} t^{3} v)

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

= lo g_{4} (4^{- 2} t^{3} v)

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

lo g_{4} (t^{3} v) : 3 lo g_{4} (t) + lo g_{4} (v)

= - 2 + (3 lo g_{4} (t) + lo g_{4} (v))

Apply rule:

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

- 2 + (3 lo g_{4} (t) + lo g_{4} (v)) = - 2 + 3 lo g_{4} (t) + lo g_{4} (v)

= - 2 + 3 lo g_{4} (t) + lo g_{4} (v)

lo g_{10} (3) + 2 lo g_{10} (4)

ln (3 x) - ln (3)

ln (4 \cdot 1 0^{14})

ln (7 e^{3})

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