## Intermediate Algebra for College Students (7th Edition)

$2 - \frac{1}{2}\log_6{(x+1)}$
RECALL: (1) $\log{(b^c)}=c \cdot \log{b}$ (2) $\log_b{(xy)} = \log_b{x} + \log_b{y}$ (3) $\log_b{(\frac{x}{y})}=\log_b{x} - \log_b{y}$ Use rule (3) above to obtain: $=\log_6{36}-\log_6{(\sqrt{x+1})}$ Note that $\sqrt{x+1} = (x+1)^{\frac{1}{2}}$ and $36=6^2$. Thus, the expression above is equivalent to: $=\log_6{(6^{2})}-\log_6{(x+1)^{\frac{1}{2}}}$ Use rule (1) above to obtain: $=2\log_6{6} - \frac{1}{2}\log_6{(x+1)}$ Use the rule $\log_b{b} = 1$ to obtain: $=2 \cdot 1 - \frac{1}{2}\log_6{(x+1)} \\=2 - \frac{1}{2}\log_6{(x+1)}$