What is log_{2}(8xy) ?

The solution to log_{2}(8xy) is 3+log_{2}(x)+log_{2}(y)

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simplify log_{2}(8xy)

simplify

lo g_{2} (8 x y)

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

Solution steps

lo g_{2} (8 x y)

Factor the number:

8 = 2^{3}

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

Apply log rule:

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

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

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

Apply log rule:

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

lo g_{2} (2^{3}) = 3

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

Apply log rule:

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

lo g_{2} (x y) = lo g_{2} (x) + lo g_{2} (y)

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

Apply rule:

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

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

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

lo g_{\frac{1}{6}} (\frac{4}{9})

- \frac{6}{8} lo g_{2} (\frac{6}{8}) - \frac{2}{8} lo g_{2} (\frac{2}{8})

- lo g_{b} (\frac{1}{5})

lo g_{10} (3) 3