Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - Chapter Test - Page 543: 24



Work Step by Step

We know that $x\log_ab=\log_ab^x$ and that $\log_ab+\log_ac=\log_a(bc)$ Hence here: $\log_4 x+\log_4(x+6)=2\\\log_4(x(x+6))=2\\x(x+6)=4^2=16\\x^2+6x-16=0\\(x+8)(x-2)=0$ Thus $x=-8$ or $x=2$. However, $x=-8$ is an extraneous solution, thus $x=2$.
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