# How do you solve log x + log (x-3) = 1?

Apr 15, 2016

$x = 5$

#### Explanation:

Assuming $\log$ is referring to the base-$10$ logarithm, we can apply the properties that $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$ and ${10}^{\log} \left(x\right) = x$ to obtain

$\log \left(x\right) + \log \left(x - 3\right) = 1$

$\implies \log \left(x \left(x - 3\right)\right) = 1$

$\implies {10}^{\log} \left(x \left(x - 3\right)\right) = {10}^{1}$

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

$\implies {x}^{2} - 3 x - 10 = 0$

$\implies x = \frac{3 \pm \sqrt{9 + 4 \cdot 10}}{2} = \frac{3}{2} \pm \frac{7}{2}$ (by the quadratic formula)

Normally we would be done, however originally we had $\log \left(x - 3\right)$ in the equation, meaning we have the restriction $x > 3$, as the logarithm function is undefined in the reals for $x \le 0$.

As $\frac{3}{2} - \frac{7}{2} < 3$ it is not a solution.

As $\frac{3}{2} + \frac{7}{2} = 5 > 3$, it is a solution.

Thus we have the solution $x = 5$