Derive that a sum equals $\log(1+x)-\log(1-x)$ Show that $$\sum_{n=1}^{\infty}\frac{x^{2n-1}}{n-1/2}=\log(1+x)-\log(1-x)$$

I got a series for $\log(1+x)$ so by replacing the $x$ by $-x$ I got that
$$\left (\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{x^n}{n}  \right )-\left ( \sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-x)^n}{n} \right )\\=\log(1+x)-\log(1-x)$$
all of this is for $|x|<1$. I do not know how to proceed with this because I should be able to rewrite the two sums into the one in the first equation or is there an easier approach I am overseeing?

First observe that:
$$\begin{aligned}
\ln(1-x)
&=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-1)^n x^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{2n+1}\frac{x^n}{n}
\\ \\ &=
-\sum_{n=1}^{\infty}\frac{x^n}{n}
\end{aligned}$$
Thus we can write the sum as:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{x^{2n-1}}{n-1/2}
&=
2\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1}
\\ \\ &=
2\left(\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{x^{2n}}{2n}\right)
\\ \\ &=
2\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{(x^2)^n}{n}
\\ \\&=
-2\ln(1-x)+\ln(1-x^2)
\\ \\ &=
\ln\left(\frac{(1-x)(1+x)}{(1-x)^2}\right)
\\ \\ &=
\ln\left(\frac{1+x}{1-x}\right)
\\ \\ &=
\ln(1+x)-\ln(1-x)
\end{aligned}$$
