Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 69


The solution is $x=9$

Work Step by Step

$\log_{3}(x-8)+\log_{3}x=2$ Combine the sum on the left side as the logarithm of a product: $\log_{3}x(x-8)=2$ Evaluate the product: $\log_{3}(x^{2}-8x)=2$ Rewrite the equation in exponential form: $x^{2}-8x=3^{2}$ $x^{2}-8x=9$ Take the $9$ to the left side: $x^{2}-8x-9=0$ Solve by factoring: $(x-9)(x+1)=0$ Set both factors equal to $0$ and solve both individual equations for $x$: $x-9=0$ $x=9$ $x+1=0$ $x=-1$ The solutions found are $x=9$ and $x=-1$. Check these solutions by plugging them into the original equation. $x=9$ $\log_{3}(9-8)+\log_{3}9=2$ $\log_{3}1+\log_{3}9=2$ $0+2=2$ $2=2$ True $x=-1$ $\log_{3}(-1-8)+\log_{3}-1=2$ False The solution is $x=9$
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