Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Review Exercises - Page 637: 26

Answer

$\left\{\dfrac{3}{2}\right\}$

Work Step by Step

Using exponents, the given equation, $ \log_2 \sqrt{8}=x $, is equivalent to \begin{align*} \log_2 \sqrt{2^3}&=x \\ \log_2 \left(2^3\right)^{1/2}&=x \\ \log_2 2^{3/2}&=x .\end{align*} Using the properties of logarithms, the equation above is equivalent to \begin{align*} \dfrac{3}{2}\log_2 2&=x &(\text{use }\log_b x^y=y\log_b x) \\\\ \dfrac{3}{2}(1)&=x \\\\ \dfrac{3}{2}&=x .\end{align*} Hence, the solution set of the equation $ \log_2 \sqrt{8}=x $ is $ \left\{\dfrac{3}{2}\right\} $.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.