# How do you solve for x in Log x + Log (3x-13)=1?

Jul 20, 2018

$x = 5$ only

#### Explanation:

If the question just says $\log$, we can safely assume that it means ${\log}_{10}$

${\log}_{10} x + {\log}_{10} \left(3 x - 13\right) = 1$

${\log}_{10} x \left(3 x - 13\right) = 1$
Recall: ${\log}_{a} b + {\log}_{a} c = {\log}_{a} b c$

$x \left(3 x - 13\right) = {10}^{1}$
Recall: If ${\log}_{a} b = c$ then ${a}^{c} = b$

$x \left(3 x - 13\right) = 10$

$3 {x}^{2} - 13 x - 10 = 0$

$\left(3 x + 2\right) \left(x - 5\right) = 0$

$x = - \frac{2}{3}$ or $x = 5$

But $x = - \frac{2}{3}$ is not a solution since you cannot log negative numbers so $x = 5$ is the solution only