Answer
Yes.
Work Step by Step
If $g$ is one-to-one then for all $y=g(x)$ there is only one $x$. Here this is satisfied, thus it is one-to-one.
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 467: 111
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