# How do we solve logsqrt(x - 8) + 1/2log(2x+ 1) = 1?

##### 1 Answer
Sep 25, 2016

We have to start by simplifying using the rule $a \log n = \log {n}^{a}$.

$\log \sqrt{x - 8} + \log \sqrt{2 x + 1} = 1$

Now, use the rule $\log n + \log m = \log \left(n \times m\right)$

$\log \left(\sqrt{x - 8} \times \sqrt{2 x + 1}\right) = 1$

$\log \left(\sqrt{2 {x}^{2} - 15 x - 8}\right) = 1$

$\sqrt{2 {x}^{2} - 15 x - 8} = {10}^{1}$

${\left(\sqrt{2 {x}^{2} - 15 x - 8}\right)}^{2} = {\left({10}^{1}\right)}^{2}$

$2 {x}^{2} - 15 x - 8 = 100$

$2 {x}^{2} - 15 x - 108 = 0$

$2 {x}^{2} - 24 x + 9 x - 108 = 0$

$2 x \left(x - 12\right) + 9 \left(x - 12\right) = 0$

$\left(2 x + 9\right) \left(x - 12\right) = 0$

$x = - \frac{9}{2} \mathmr{and} 12$

However, $x = - \frac{9}{2}$ is extraneous since it renders the square root negative, which is undefined in the real number system. Hence, the solution set is $\left\{12\right\}$.

Hopefully this helps!