What is 1/9 (log_{2}(z)+4log_{2}(y)) ?

The solution to 1/9 (log_{2}(z)+4log_{2}(y)) is (log_{2}(z)+4log_{2}(y))/9

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simplify 1/9 (log_{2}(z)+4log_{2}(y))

simplify

\frac{1}{9} (lo g_{2} (z) + 4 lo g_{2} (y))

\frac{lo g _{2} ( z ) + 4 lo g _{2} ( y )}{9}

Solution steps

\frac{1}{9} (lo g_{2} (z) + 4 lo g_{2} (y))

Apply the distributive law:

a (b + c) = ab + a c

\frac{1}{9} (lo g_{2} (z) + 4 lo g_{2} (y)) = \frac{1}{9} lo g_{2} (z) + \frac{1}{9} \cdot 4 lo g_{2} (y)

= \frac{1}{9} lo g_{2} (z) + \frac{1}{9} \cdot 4 lo g_{2} (y)

Simplify

\frac{1}{9} lo g_{2} (z) + \frac{1}{9} \cdot 4 lo g_{2} (y) : \frac{lo g _{2} ( z ) + 4 lo g _{2} ( y )}{9}

= \frac{lo g _{2} ( z ) + 4 lo g _{2} ( y )}{9}

\frac{3 lo g _{10} ( 2 )}{lo g _{10} ( 2 ) - lo g _{10} ( 3 )}

lo g_{10} (z^{6} y)

lo g_{2} (3 x y^{2})

lo g_{10} (8 k)

ln ((x y)^{6})

What is 1/9 (log_{2}(z)+4log_{2}(y)) ?
The solution to 1/9 (log_{2}(z)+4log_{2}(y)) is (log_{2}(z)+4log_{2}(y))/9