College Algebra (11th Edition)

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

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 446: 12

Answer

$x\approx0.823$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2^{x+3}=5^{2x} ,$ take the logarithm of both sides. Then use the properties of logarithms and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log2^{x+3}=\log5^{2x} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x+3)\log2=2x\log5 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(\log2)+3(\log2)=2x\log5 \\\\ x\log2+3\log2=2x\log5 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x\log2-2x\log5=-3\log2 \\\\ x(\log2-2\log5)=-3\log2 \\\\ x=-\dfrac{3\log2}{\log2-2\log5} \\\\ x\approx0.823 .\end{array}
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