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

Jan 6, 2016

$x = 5$

Explanation:

$\log x + \log \left(x - 3\right) = 1$
We know that: $\log a + \log b = \log \left(a \cdot b\right)$
$\implies L o g \left(x \left(x - 3\right)\right) = 1$
$\implies \log \left({x}^{2} - 3 x\right) = 1$
$\implies {x}^{2} - 3 x = 10$
$\implies {x}^{2} - 3 x - 10 = 0$
$\implies \left(x - 5\right) \left(x + 2\right) = 0$
$\implies x = 5 , - 2$

Verification:-
Put $x = 5$
$L . H . S = L o g x + L o g \left(x - 3\right) = L o g 5 + \log \left(5 - 3\right) = \log 5 + \log 2 = \log \left(5 \cdot 2\right) = \log 10 = 1 = R . H . S$
Verified.
Put $x = - 2$
$L . H . S = L o g x + L o g \left(x - 3\right) = L o g \left(- 2\right) + \log \left(- 2 - 3\right) = \log \left(- 2\right) + \log \left(- 5\right)$
Here we have to find the log of a negative number which is undefined.
Therefore not verified.

Therefore, only $x = 5$ is true.