Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - 7-4 Properties of Logarithms - Practice and Problem-Solving Exercises - Page 468: 89

Answer

$w=2$

Work Step by Step

In order to get rid of the square root, square both sides: \begin{align*} (2\sqrt{w-1})^2&=(\sqrt{w+2})^2\\ 2^2(w-1)&=w+2\\ 4(w-1)&=w+2\\ 4w-4&=w+2 \end{align*} Move the $w$ terms to one side: $4w-w=2+4$ $3w=6$ $w=6/3$ $w=2$ Test the solution to make sure that it is not extraneous: $2\sqrt{2-1}=\sqrt{2+2}$ $2\sqrt{1}=\sqrt{4}$ $2\cdot 1=2$ $2=2$ The solution works.
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.