Answer
$x=2$
Work Step by Step
$$log(x + 3) + log(x) = 1$$
By rules of logarithmic functions:
$log(x + 3) + log(x) = log(x)(x + 3) = log (x^{2} + 3x)$
Therefore, we can re-write the original equation as follows and solve accordingly:
$log_{10}(x^{2} + 3x) = 1$
$10^{1} = (x^{2} + 3x)$
$0 = x^{2} + 3x - 10$; where 2 factors of 10 that,when substracted result in +3 are $5$ and $2$:
$0=(x + 5)(x-2)$; where $x=-5$ or $x=2$. Since logarithms are not defined for negative values within the realm of Real numbers, the only solution is $x = 2$.
