What is taylor 5^t,0 ?

The answer to taylor 5^t,0 is 1+ln(5)t+(ln^2(5))/2 t^2+(ln^3(5))/6 t^3+(ln^4(5))/(24)t^4+\ldots

taylor 5^t,0

taylor

5^{t}, 0

1 + ln (5) t + \frac{ln ^{2} ( 5 )}{2} t^{2} + \frac{ln ^{3} ( 5 )}{6} t^{3} + \frac{ln ^{4} ( 5 )}{24} t^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{ln ( 5 )}{1 !} t + \frac{ln ^{2} ( 5 )}{2 !} t^{2} + \frac{ln ^{3} ( 5 )}{3 !} t^{3} + \frac{ln ^{4} ( 5 )}{4 !} t^{4} + \dots

Refine

= 1 + ln (5) t + \frac{ln ^{2} ( 5 )}{2} t^{2} + \frac{ln ^{3} ( 5 )}{6} t^{3} + \frac{ln ^{4} ( 5 )}{24} t^{4} + \dots

\int_{0}^{1} (u + 7) (u - 8) d u

\int 2 e^{4 x} d x

\frac{5}{3 ( s + 3 ) ^{2} + 27}

x \to - 1 lim (\frac{x}{x - 1})

f (x) = - x^{2} + 6 x - 8, 4, - 2

What is taylor 5^t,0 ?
The answer to taylor 5^t,0 is 1+ln(5)t+(ln^2(5))/2 t^2+(ln^3(5))/6 t^3+(ln^4(5))/(24)t^4+\ldots