Use laws of logarithms to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ I am to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ using the laws of logarithms. The solution is provided as $\ln(2)+\ln(b)+\frac{\ln(b+1)-\ln(b-1)}{2}$ whereas I got $\ln(2)+\ln(b)+\frac{1}{2}\ln(b+1)-\frac{1}{2}\ln(b-1)$
My working:
$$\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$$
$$=\ln(2)+\ln(b)+\ln\left(\sqrt{\frac{b+1}{b-1}}\right)$$
$$=\ln(2)+\ln(b)+\frac{1}{2}\ln\left(\frac{b+1}{b-1}\right)$$
$$=\ln(2)+\ln(b)+\frac{1}{2}\ln(b+1)-\frac{1}{2}\ln(b-1).$$
I suspect that I might have made a mistake using the power rule to use $1/2$ for the root of the expression $\frac{b+1}{b-1}$, but I'm unsure.
How can I arrive at the correct solution?

$$\frac{1}{2}A-\frac{1}{2}B=\frac{A-B}{2}$$
where $A=\ln(b+1)$, $B=\ln(b-1)$.
In short, you arrived at the given answer.
