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: 96


The solution set is $x=$ {$\frac{1}{100}, 100$}

Work Step by Step

$$3|\log x|-6=0$$ *Domain: $x\in R, x\gt0$ $$3|\log x|=6$$ $$|\log x|=2$$ As $|X|=A$, either $X=A$ or $X=-A$. Therefore, $$\log x=2$$ or $$\log x=-2$$ * For $\log x=2$ $$\log x=2$$ $$x=10^2=100$$ - Check again: $$3|\log100|-6=3|2|-6=3\times2-6=0$$ That means $x=100$ is a solution to this problem. * For $\log x=-2$ $$\log x=-2$$ $$x=10^{-2}$$ $$x=\frac{1}{10^2}=\frac{1}{100}$$ - Check again: $$3|\log\frac{1}{100}|-6=3|\log10^{-2}|-6=3|-2|-6=3\times2-6=0$$ $x=\frac{1}{100}$ is also a solution to this problem. In conclusion, the solution set is $x=$ {$\frac{1}{100}, 100$}
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