logarithm equation with scalar on right hand side $\log_7 (x^2-1) - \log_7 (x-1) = 2$
$\log_7 49 = 2$
=> $\log_7 (x^2-1) - \log_7 (x-1) = \log_7 49$
=> $\frac{(x^2-1)}{x-1} = 49$
=> $x^2 -1 = 49(x -1)$
=> $x^2 -1 = 49x -49$
=> $x^2 - 49x + 48 = 0$
The answer in the book is 48 so I'm obviously doing this wrong.

We have that $x>1$ ($\log_7x$ is only defined if $x>0$) and
$$
\log_7(x^2-1)-\log_7(x-1)=\log_7\biggl(\frac{x^2-1}{x-1}\biggr)=\log_7\biggl(\frac{(x-1)(x+1)}{x-1}\biggr)=\log_7(x+1)=2.
$$
Hence, $x+1=49$ and $x=48$.
