Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - Chapter Review - Chapter Test - Page 348: 15

Answer

$\{91\}$

Work Step by Step

We know that if $\log{y}=x$ then $10^x=y$. Hence, $\log{(x+9)}=2 \longrightarrow 10^2=x+9$ Solve the equarton above to obtain $100=x+9\\ 100-9=x+9-9\\ 91=x$ The solution set is: $\{91\}.$
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