Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 813: 80



Work Step by Step

Since, we have two equations $2 \log x=y+3$ and $\log x=y-1$ Thus, $2 (y-1)=y+3$ or, $y=5$ Now, $\log x=y-1$ or, $\log x=5-1 \implies 10^4=x$ and $x=10000$ Hence, $(x,y)=${$10000,5$}
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