College Algebra (11th Edition)

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

Chapter 4 - Review Exercises: 38

Answer

$x\approx2.269$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2^{x+3}=5^x ,$ take the logarithm of both sides. Then use the laws of logarithms to isolate the variable. Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log2^{x+3}=\log5^x .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the equation above is equivalent to \begin{array}{l}\require{cancel} (x+3)\log2=x\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)=x\log5 \\\\ x\log2+3\log2=x\log5 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x\log2-x\log5=-3\log2 \\\\ x(\log2-\log5)=-3\log2 \\\\ x=-\dfrac{3\log2}{\log2-\log5} \\\\ x\approx2.269 .\end{array}
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