What is taylor ln(4-3x) ?

The answer to taylor ln(4-3x) is 2ln(2)-3/4 x-9/32 x^2-9/64 x^3-81/1024 x^4+\ldots

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taylor ln(4-3x)

taylor

ln (4 - 3 x)

2 ln (2) - \frac{3}{4} x - \frac{9}{32} x^{2} - \frac{9}{64} x^{3} - \frac{81}{1024} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 2 ln (2) + \frac{\frac{d}{d x} ( ln ( 4 - 3 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 4 - 3 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 4 - 3 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 2 ln (2) + \frac{- \frac{3}{4}}{1 !} x + \frac{- \frac{9}{16}}{2 !} x^{2} + \frac{- \frac{27}{32}}{3 !} x^{3} + \frac{- \frac{243}{128}}{4 !} x^{4} + \dots

Refine

= 2 ln (2) - \frac{3}{4} x - \frac{9}{32} x^{2} - \frac{9}{64} x^{3} - \frac{81}{1024} x^{4} + \dots

x y^{2} y^{^{'}} = x + 3

\int x^{3} ln (x^{3}) d x

y^{^{'}} = \frac{1}{sec ^{2} ( y )}

\frac{d}{d x} (\frac{sin ( 3 x )}{cos ( 3 x )})

x \to 2 lim (e^{2 x - x^{2}})

What is taylor ln(4-3x) ?
The answer to taylor ln(4-3x) is 2ln(2)-3/4 x-9/32 x^2-9/64 x^3-81/1024 x^4+\ldots