## College Algebra (6th Edition)

$$x = 10^{\frac{3}{2}}$$ $$OR$$ $$x = \frac{1}{100}$$
$$\log(x) (2\log(x) + 1) = 6$$ Let $a = \log(x)$, and continue solving: $$a(2a + 1) = 6$$ $$2a^2 + a – 6 = 0$$ $$a = \frac{-(1) \frac{+}{} \sqrt{(1)^2 – 4(2)(-6)}}{2(2)}$$ $$a = \frac{ -1 \frac{+}{} \sqrt{1 + 48}}{4}$$ $$a = \frac{-1 \frac{+}{} 7}{4}$$ We substitute the value of $a$ to its original form: $$\log x = \frac{-1 \frac{+}{} 7}{4}$$ $$x = 10^{\frac{-1 \frac{+}{} 7}{4}}$$ $$x = 10^{\frac{6}{4}} = 10^{\frac{3}{2}}$$ $$OR$$ $$x = 10^{-2} = \frac{1}{100}$$