College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises: 51

Answer

$x\approx6.959$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 4(1.06)^x+2=8 ,$ use the properties of equality to isolate the exponential expression. Then take the logarithm of both sides and use the properties of logarithms and of equality to isolate the variable. Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the exponential expression results to \begin{array}{l}\require{cancel} 4(1.06)^x=8-2 \\\\ 4(1.06)^x=6 \\\\ (1.06)^x=\dfrac{6}{4} \\\\ (1.06)^x=\dfrac{3}{2} .\end{array} Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log(1.06)^x=\log\dfrac{3}{2} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log(1.06)^x=\log3-\log2 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} x\log(1.06)=\log3-\log2 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x=\dfrac{\log3-\log2}{\log(1.06)} \\\\ x\approx6.959 .\end{array}
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