What is expand (ln(a^4b^{-1}))/(ln((bc)-4)) ?

The solution to expand (ln(a^4b^{-1}))/(ln((bc)-4)) is (4ln(a))/(ln(bc-4))-(ln(b))/(ln(bc-4))

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expand (ln(a^4b^{-1}))/(ln((bc)-4))

expand

\frac{ln ( a ^{4} b ^{- 1} )}{ln ( ( b c ) - 4 )}

\frac{4 ln ( a )}{ln ( b c - 4 )} - \frac{ln ( b )}{ln ( b c - 4 )}

Solution steps

\frac{ln ( a ^{4} b ^{- 1} )}{ln ( ( b c ) - 4 )}

Remove parentheses:

(a) = a

= \frac{ln ( a ^{4} b ^{- 1} )}{ln ( b c - 4 )}

Multiply

a^{4} b^{- 1} : \frac{a ^{4}}{b}

= \frac{ln ( \frac{a ^{4}}{b} )}{ln ( b c - 4 )}

Simplify

ln (\frac{a ^{4}}{b}) : 4 ln (a) - ln (b)

= \frac{4 ln ( a ) - ln ( b )}{ln ( b c - 4 )}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 ln ( a ) - ln ( b )}{ln ( b c - 4 )} = \frac{4 ln ( a )}{ln ( b c - 4 )} - \frac{ln ( b )}{ln ( b c - 4 )}

= \frac{4 ln ( a )}{ln ( b c - 4 )} - \frac{ln ( b )}{ln ( b c - 4 )}

(x + 2 x^{2} - x) (x - 2 x^{2} - x)

\frac{0.00550368}{s ( 0.00224 + s ^{2} + 0.00182 s )}

(y e^{x y} - \frac{1}{y ^{2}}) \cdot (e^{x^{2}} \cdot 2 x + 4 x)

(- \frac{x ^{2}}{3} - 1) (2 x - 2)

- \frac{u ( 16 x ^{6} - 14 u - 7 u x ^{2} )}{2 x}

What is expand (ln(a^4b^{-1}))/(ln((bc)-4)) ?
The solution to expand (ln(a^4b^{-1}))/(ln((bc)-4)) is (4ln(a))/(ln(bc-4))-(ln(b))/(ln(bc-4))