Answer
$2 - \frac{1}{2}\log_8{(x+1)}$
Work Step by Step
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:
$=\log_8{64}-\log_8{(\sqrt{x+1})}$
Note that $\sqrt{x+1} = (x+1)^{\frac{1}{2}}$ and $64=8^2$.
Thus, the expression above is equivalent to:
$=\log_8{(8^{2})}-\log_8{(x+1)^{\frac{1}{2}}}$
Use rule (1) above to obtain:
$=2\log_8{8} - \frac{1}{2}\log_8{(x+1)}$
Use the rule $\log_b{b} = 1$ to obtain:
$=2 \cdot 1 - \frac{1}{2}\log_8{(x+1)}
\\=2 - \frac{1}{2}\log_8{(x+1)}$
