What is expand log_{4}((4a)/(b^3)) ?

The solution to expand log_{4}((4a)/(b^3)) is 1+log_{4}(a)-3log_{4}(b)

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expand log_{4}((4a)/(b^3))

expand

lo g_{4} (\frac{4 a}{b ^{3}})

1 + lo g_{4} (a) - 3 lo g_{4} (b)

Solution steps

lo g_{4} (\frac{4 a}{b ^{3}})

Apply log rule:

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

lo g_{4} (\frac{4 a}{b ^{3}}) = lo g_{4} (4 a) - lo g_{4} (b^{3})

= lo g_{4} (4 a) - lo g_{4} (b^{3})

Apply log rule:

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

lo g_{4} (b^{3}) = 3 lo g_{4} (b)

= lo g_{4} (4 a) - 3 lo g_{4} (b)

Apply log rule:

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

lo g_{4} (4 a) = lo g_{4} (4) + lo g_{4} (a)

= lo g_{4} (4) + lo g_{4} (a) - 3 lo g_{4} (b)

Apply log rule:

lo g_{a} (a) = 1

= 1 + lo g_{4} (a) - 3 lo g_{4} (b)

\frac{1}{3} lo g_{10} (2)

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

\frac{1}{2} ln (\frac{1 + r}{1 - r})

ln (\frac{1}{β ^{n}} e^{\frac{- 1}{β ^{n}} (- α n + x)})

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

What is expand log_{4}((4a)/(b^3)) ?
The solution to expand log_{4}((4a)/(b^3)) is 1+log_{4}(a)-3log_{4}(b)