## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Review Exercises: 64

#### Answer

$x=3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log_{16}\sqrt{x+1}=\dfrac{1}{4} ,$ change to exponential form. Use the definition of rational exponents to simplify the result. Then square both sides and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \sqrt{x+1}=16^{\frac{1}{4}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt{x+1}=\sqrt[4]{16} \\\\ \sqrt{x+1}=\sqrt[4]{2^4} \\\\ \sqrt{x+1}=2 .\end{array} Squaring both sides and using the properties of equality to isolate the variable result to \begin{array}{l}\require{cancel} (\sqrt{x+1})^2=2^2 \\\\ x+1=4 \\\\ x=4-1 \\\\ x=3 .\end{array}

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