If a and b are the same sign but different then prove: $\log|a+b|-\log|a-b|=\frac{1}{2}\log2 <>a^2+b^2=6ab$

Question

If a and b are the same sign but different then prove: $$\log|a+b|-\log|a-b|=\frac{1}{2}\log2 \lt \gt a^2+b^2=6ab$$

Answer 1

From $a^2+b^2=6ab$ we have $(a+b)^2=8ab$ and $(a-b)^2=4ab$ so \begin{eqnarray*} \ln(a+b)-\ln(a-b)= \frac{1}{2} (\ln(8)-\ln(4)) =\frac{\ln(2)}{2}. \end{eqnarray*}