College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 41

Answer

$x\approx2.79$

Work Step by Step

Find the natural log of both sides and then pull out the exponents on both sides. Then solve for $x$. $5^{2x+3}=3^{x-1}$ $(2x+3)\ln(5)=(x-1)\ln(3)$ $\frac{2x+3}{x-1}=\frac{\ln(3)}{\ln(5)}$ $\frac{2x+3}{x-1}=0.68$ $0.68(x-1)=2x+3$ $0.68x-0.68=2x+3$ $0.68x=2x+3.68$ $1.32x=3.68$ $x\approx2.79$
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