What is expand (i-3)ln(-4i+2) ?

The solution to expand (i-3)ln(-4i+2) is iln(-4i+2)-3ln(-4i+2)

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expand (i-3)ln(-4i+2)

expand

(i - 3) ln (- 4 i + 2)

i ln (- 4 i + 2) - 3 ln (- 4 i + 2)

Solution steps

(i - 3) ln (- 4 i + 2)

= ln (- 4 i + 2) (i - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = ln (- 4 i + 2), b = i, c = 3

= ln (- 4 i + 2) i - ln (- 4 i + 2) \cdot 3

= i ln (- 4 i + 2) - 3 ln (- 4 i + 2)

\frac{2}{( z - 2 j ) ( z + j )}

4 d (2 d + 3 e)

(\frac{v ^{1} - v}{a})^{2}

\frac{1}{5} (3 + ln (18))

\frac{( q + 1 ) ( t + 2 )}{( t + 1 )}

What is expand (i-3)ln(-4i+2) ?
The solution to expand (i-3)ln(-4i+2) is iln(-4i+2)-3ln(-4i+2)