Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=log_ax$, we have $LHS=-f(x)=-log_ax=-\frac{logx}{loga}$
Step 2. $RHS=log_{1/a}x=\frac{logx}{log(1/a)}=\frac{logx}{-log(a)}$
Step 3. We can see that $LHS=RHS$.
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 306: 105
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