# How do you solve  1/2log_a(x + 2) + 1/2log_a(x - 1) = 2/3log_a27?

Dec 4, 2015

I found: $x = \frac{- 1 + 3 \sqrt{37}}{2}$

#### Explanation:

We can try some manipulations:
collect $\frac{1}{2}$:
$\frac{1}{2} \left[{\log}_{a} \left(x + 2\right) + {\log}_{a} \left(x - 1\right)\right] = \frac{2}{3} {\log}_{a} 27$
${\log}_{a} \left(x + 2\right) + {\log}_{a} \left(x - 1\right) = 2 \cdot \frac{2}{3} {\log}_{a} 27$

use the property of logs that says:
$\log x + \log y = \log \left(x y\right)$
${\log}_{a} \left[\left(x + 2\right) \left(x - 1\right)\right] = \frac{4}{3} {\log}_{a} 27$

use the other property of logs that says:
$a \log x = \log {x}^{a}$
${\log}_{a} \left[\left(x + 2\right) \left(x - 1\right)\right] = {\log}_{a} {27}^{\frac{4}{3}}$

if the two logs must be equal then the argumens must be as well:
$\left(x + 2\right) \left(x - 1\right) = {27}^{\frac{4}{3}}$
$\left(x + 2\right) \left(x - 1\right) = \sqrt{{27}^{4}}$
$\left(x + 2\right) \left(x - 1\right) = 81$
${x}^{2} - x + 2 x - 2 - 81 = 0$
${x}^{2} + x - 83 = 0$

You can now use the Quadratic Formula to get:
${x}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 332}}{2} =$
two solutions:
${x}_{1} = \frac{- 1 - 3 \sqrt{37}}{2}$ NO it will give you a negative log.
${x}_{2} = \frac{- 1 + 3 \sqrt{37}}{2}$