Intermediate Algebra (6th Edition)

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

Chapter 9 - Review: 62


$x=1$ and $x=-8$

Work Step by Step

If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $log_{8}(x^{2}+7x)=1$ means $8^{1}=8=x^{2}+7x$. Subtract 8 from both sides of the equation to get all terms on one side. $x^{2}+7x-8=0$ We know that 1 and 8 are factors of 8 (and the sum of -1 and 8 is 7, which is the coefficient attached to the middle term). Therefore, we can factor this equation into $(x-1)(x+8)=0$. After setting both terms equal to 0, we know that $x=1$ and $x=-8$.
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