Answer
$16$
Work Step by Step
We know that $k\log_{a^k}M=\log_aM$, hence $\log_{16}x+\log_4x+\log_2x=\log_{2^4}x+\log_{2^2}x+\log_{2^1}x=\log_{16}x+2\log_{16}x+4\log_{16}x=7\log_{16}x=7\\log_{16}x=1\\\\x=16^1=16$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Section 6.6 - Logarithmic and Exponential Equations - 6.6 Assess Your Understanding - Page 466: 85
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