Is the equality below true over all complex numbers? $$\log\left(\frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}\right)=3\log\left(\frac{1+x}{1-x}\right)$$
The L.H.S. (Left hand side expression) simplifies to $\log\left(-\frac{(x+1)^3}{(x-1)^3}\right)=\log\left(-\left(\frac{x+1}{x-1}\right)^3\right)$
But I think you cannot take the exponent $3$ outside the $\log$ because of that negative sign; if you are allowed to, then yes, the $(x-1)$ becomes $(1-x)$ as desired, but I want to confirm.
So essentially, is the below equality true? $$\log\left(-\left(\frac{x+1}{x-1}\right)^3\right)=3\log\left(\frac{1+x}{1-x}\right)$$
$$\log(-a^3)=\log((-a)^3)=3\log(-a).$$
In complex numbers, the property doesn't necessarily hold because of the indeterminacy of the imaginary part. For instance, with real $a>0$ and choosing the principal branch
$$\log(-a^3)=\log(|a|^3)+i\pi\ne3\log(-a)=3(\log(|a|)+i\pi).$$