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.

Since $x=1$ is excluded, your equation
$$
\frac{(x^2-1)}{x-1} = 49
$$ is just
$$
x+1=49.
$$
