Chapter 7: Problem 142
$$ 2 \log _{2}\left(\frac{x-7}{x-1}\right)+\log _{2}\left(\frac{x-1}{x+1}\right)=1 $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 7: Problem 142
$$ 2 \log _{2}\left(\frac{x-7}{x-1}\right)+\log _{2}\left(\frac{x-1}{x+1}\right)=1 $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
$$ \sqrt[3]{x+1}+\sqrt[3]{3 x+1}=\sqrt[3]{x-1} $$
Show that the polynomial \(P(x)=x^{7}+x^{5}-2 x^{4}+x^{3}-3 x^{2}+7 x-5\) cannot have a negative real root.
$$ 3^{4 x+8}-4 \cdot 3^{2 x+5}+28=2 \log _{2} \sqrt{2} $$
$$ \sqrt{25-x}=2-\sqrt{9+x} $$
$$ \left(x^{2}-4\right) \sqrt{x+1}=0 $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.