Answer
$1+\frac{1}{2}\log_8{3}+\frac{5}{2}\log_8{a}$
Work Step by Step
Recall the product property of logarithms (pg. 462):
$\log_b{mn}=\log_b{m}+\log_b{n}$
Applying the product property to the given expression gives::
$\log_8{8\sqrt{3a^5}}=\log_8{8}+\log_8{\sqrt{3a^5}}$
Next, recall the power property of logarithms (pg. 462):
$\log_b{m^n}=n\log_b{m}$
Applying the power property, we get:
$\log_8{8}+\log_8{\sqrt{3a^5}}=\log_8{8}+\log_8{(3a^5)^{1/2}}=\log_8{8}+\frac{1}{2}\log_8{(3a^5)}$
Next, apply the product property again to expand further:
$\log_8{8}+\frac{1}{2}\log_8{(3a^5)}=\log_8{8}+\frac{1}{2}\log_8{3}+\frac{1}{2}\log_8{a^5}$
We apply the power property once more to simplify:
$\log_8{8}+\frac{1}{2}\log_8{3}+\frac{1}{2}\log_8{a^5}=\log_8{8}+\frac{1}{2}\log_8{3}+\frac{5}{2}\log_8{a}$
Lastly, we note that $\log_8{8}=1$ (because $8^1=8$).
$\log_8{8}+\frac{1}{2}\log_8{3}+\frac{5}{2}\log_8{a}=1+\frac{1}{2}\log_8{3}+\frac{5}{2}\log_8{a}$
