## Precalculus (6th Edition) Blitzer

$\displaystyle \log\frac{x(x-1)}{7}$
$\log x+\log(x^{2}-1)-\log 7-\log(x+1)=$ ... group all positive signed terms; group all negative signed terms $=[\log x+\log(x^{2}-1)]-[\log 7+\log(x+1)]$ ... apply the product rule to each bracket $=\log[x(x^{2}-1)]-\log[7(x+1)]$ ... apply the quotient rule $=\displaystyle \log\frac{x(x^{2}-1)}{7(x+1)}$ ... recognize a difference of squares in the numerator $=\displaystyle \log\frac{x(x+1)(x-1)}{7(x+1)}$ ... cancel the common term $=\displaystyle \log\frac{x(x-1)}{7}$