Answer
$1$
Work Step by Step
Recall:
$\log _{a}m+\log _{a}n=\log _{a}\left( m\times n\right)$
Thus,
$\log _{8}2+\log _{8}4=\log _{8}\left( 2\times 4\right) =\log _{8}8$
Since $\log_a{a}=1$, then
$\log_8{8}=1$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 19
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