College Algebra (11th Edition)

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

Chapter 4 - Review Exercises: 44

Answer

$x\approx-17.531$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 5^{x+2}=2^{2x-1} ,$ take the logarithm of both sides. Use the laws of logarithms and the properties of equality to isolate the variable. Finally, Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the natural logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log5^{x+2}=\log2^{2x-1} .\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+2)\log5=(2x-1)\log2 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(\log5)+2(\log5)=2x(\log2)-1(\log2) \\\\ x\log5+2\log5=2x\log2-\log2 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} \log2+2\log5=2x\log2-x\log5 \\\\ \log2+2\log5=x(2\log2-\log5) \\\\ \dfrac{\log2+2\log5}{2\log2-\log5}=x \\\\ x=\dfrac{\log2+2\log5}{2\log2-\log5} \\\\ x\approx-17.531 .\end{array}
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