Determining the Taylor Series of $\log_3(2x-1)$

I know the pattern of the Taylor series but I do not know how to formulate the formula.

The given is $$f(x) = \log_3(2x-1)$$ at $$x = 1$$.

$$f^{(0)}(x) = \log_3(2x-1)$$

$$f^{(1)}(x) = \frac{1}{\log3} \frac{2}{(2x-1)}$$

$$f^{(2)}(x) = \frac{1}{\log3} \frac{-4}{(2x-1)^2}$$

$$f^{(3)}(x) = \frac{1}{\log3} \frac{16}{(2x-1)^3}$$

$$f^{(4)}(x) = \frac{1}{\log3} \frac{-96}{(2x-1)^4}$$

$$f^{(0)}(1) = \log_3(1)$$

$$f^{(1)}(1) = \frac{1}{\log3} \cdot 2$$

$$f^{(2)}(1) = \frac{1}{\log3} \cdot (-4)$$

$$f^{(3)}(1) = \frac{1}{\log3} \cdot 16$$

$$f^{(4)}(1) = \frac{1}{\log3} \cdot (-96)$$

I have a problem on how to formulate the numerator part of the Taylor series. I am uncertain on what to do with pattern $$2, -4, 16, -96,\ldots$$.

• Around $x=$ what ? Jun 9 '21 at 9:23
• If the Taylor expansion should be done around $1$, just write $x=1+h$, express $\log_3$ in terms of $\log=\ln$, and use the Taylor expansion formula for $-\ln(1-y)$, obtained e.g. by using the one for its derivative. Jun 9 '21 at 10:25
• Hint: $$\log _3 (2x - 1) = \frac{1}{{\log 3}}\log (2x - 1) = \frac{1}{{\log 3}}\log (1 + 2(x - 1)).$$
– Gary
Jun 9 '21 at 11:07

I think the doubt is how to generalize the Taylor expansion of the given function around a point $$x$$. We note that every time we differentiate a function, the term will be multiplied by $$2$$ (coming from $$2x-1$$), and alternatively, $$-1$$ gets multiplied. We will also get a factorial term coming from the power of the term $$(2x-1)$$. Thus, except for the first term, we can write $$f^{(n)}(x)=\frac{(-1)^{n+1}}{log(3)}\frac{2^n}{(2x-1)^n}(n-1)!$$ Hope this helps.