Answer
See below.
Work Step by Step
We know that $\log_a x^n=n\cdot \log_a x$, hence $\log_2 {(1+1)}\log_22^2=1$, whereas $\log_2 1+\log_21=\log_22^0+\log_22^0=0+0=0$ and $0\ne1$
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 6 - Section 6.5 - Properties of Logarithms - 6.5 Assess Your Understanding - Page 460: 113
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