What is expand log_{2}((8w)/(g^3)) ?

The solution to expand log_{2}((8w)/(g^3)) is 3+log_{2}(w)-3log_{2}(g)

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expand log_{2}((8w)/(g^3))

expand

lo g_{2} (\frac{8 w}{g ^{3}})

3 + lo g_{2} (w) - 3 lo g_{2} (g)

Solution steps

lo g_{2} (\frac{8 w}{g ^{3}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

lo g_{2} (\frac{8 w}{g ^{3}}) = lo g_{2} (8 w) - lo g_{2} (g^{3})

= lo g_{2} (8 w) - lo g_{2} (g^{3})

Apply log rule:

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

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

= lo g_{2} (8 w) - 3 lo g_{2} (g)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{2} (8 w) = lo g_{2} (8) + lo g_{2} (w)

= lo g_{2} (8) + lo g_{2} (w) - 3 lo g_{2} (g)

lo g_{2} (8) = 3

= 3 + lo g_{2} (w) - 3 lo g_{2} (g)

ln (\frac{6.61}{1.74}) \cdot 8.314

ln (5 + x^{2}) + \frac{1}{9} ln (x) - ln (x + 1)

lo g_{4} (x) + 6 lo g_{4} (y) + 3 lo g_{4} (z)

\frac{1}{2} (ln (e^{2} + 1) - ln (2))

lo g_{6} (\frac{6}{x})

What is expand log_{2}((8w)/(g^3)) ?
The solution to expand log_{2}((8w)/(g^3)) is 3+log_{2}(w)-3log_{2}(g)