Question

Solve for x: logbase3 (2x+1)-logbase3 (x-1)-1=0

Answer 1

\[\log_3(2x+1)-\log_3 (x-1)-1=0\]\[\log_3\frac{(2x+1)}{(x-1)} -1=0\]\[\log_3\frac{(2x+1)}{(x-1)} =1\]\[\log_3\frac{(2x+1)}{(x-1)} =1\times \log_33\]\[\log_3\frac{(2x+1)}{(x-1)} =\log_33\]\[\frac{(2x+1)}{(x-1)} =3\]\[(2x+1)=3(x-1)\]\[2x+1=3x-3\]\[4=x\]