Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Section 9.8 - Exponential and Logarithmic Equations and Problem Solving - Exercise Set - Page 590: 21

Answer

$x=\dfrac{3}{24}$

Work Step by Step

Apply the quotient rule to have: $\log_5{\left(\frac{x+3}{x}\right)}=2$ RECALL: $\log_a{b}=x \longrightarrow a^x=b $ where $b\gt0$ Use this rule in converting the equation to its equivalent exponential equation to have: $5^2=\dfrac{x+3}{x} \\25=\dfrac{x+3}{x} \\25(x)=x+3 \\25x=x+3 \\25x-x=3 \\24x=3 \\x=\dfrac{3}{24}$
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