# What is the solution set of log_7 (x+3) + log_ 7 (x-5) = 1?

Mar 22, 2018

$x = 1 + \sqrt{23} \approx 5.7958$

#### Explanation:

${\log}_{7} \left(x + 3\right) + {\log}_{7} \left(x - 5\right) = 1$

Applying ${\log}_{n} a + {\log}_{n} b = {\log}_{n} a b$

${\log}_{7} \left(x + 3\right) \left(x - 5\right) = 1$

Now, $1 = {\log}_{7} 7$

$\therefore {\log}_{7} \left(x + 3\right) \left(x - 5\right) = {\log}_{7} 7$

If ${\log}_{n} a = {\log}_{n} b$ then $a = b$

Thus, $\left(x + 3\right) \left(x - 5\right) = 7$

${x}^{2} - 2 x - 15 = 7$

${x}^{2} - 2 x - 22 = 0$

$x = \frac{2 \pm \sqrt{4 + 4 \times 22}}{2}$

$= 1 \pm \frac{\sqrt{92}}{2} = 1 \pm \frac{2 \sqrt{23}}{2}$

$= 1 \pm \sqrt{23}$

Remember that ${\log}_{n} \left(x\right)$ is undefined for $x < 0$

Then, since $\sqrt{23} > 1$, we can reject the negative result.

NB: since $1 + \sqrt{23} > 5$, we can retain the positive result

$\therefore x = 1 + \sqrt{23} \approx 5.7958$