What is taylor ln(5-2x) ?

The answer to taylor ln(5-2x) is ln(5)-2/5 x-2/25 x^2-8/375 x^3-4/625 x^4+\ldots

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taylor ln(5-2x)

taylor

ln (5 - 2 x)

ln (5) - \frac{2}{5} x - \frac{2}{25} x^{2} - \frac{8}{375} x^{3} - \frac{4}{625} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= ln (5) + \frac{\frac{d}{d x} ( ln ( 5 - 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 5 - 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 5 - 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= ln (5) + \frac{- \frac{2}{5}}{1 !} x + \frac{- \frac{4}{25}}{2 !} x^{2} + \frac{- \frac{16}{125}}{3 !} x^{3} + \frac{- \frac{96}{625}}{4 !} x^{4} + \dots

Refine

= ln (5) - \frac{2}{5} x - \frac{2}{25} x^{2} - \frac{8}{375} x^{3} - \frac{4}{625} x^{4} + \dots

x \to \infty + lim (x^{12})

\frac{d}{d t} (sin (\frac{t}{2}))

y^{^{'}} = 5

\int e^{\frac{3}{2} x} d x

t \frac{d y}{d t} + y = t^{2}

What is taylor ln(5-2x) ?
The answer to taylor ln(5-2x) is ln(5)-2/5 x-2/25 x^2-8/375 x^3-4/625 x^4+\ldots