Answer
$x = 1$
Work Step by Step
$$log_{3}(x) + log_{3}(x + 2) = 1$$ Since both logarithms have the same base, we can use the Rules of Logarithms to simplify the logarithmic terms as follows: $$log_{3}(x) + log_{3}(x + 2)$$ $$log_{3}(x)(x+2)$$ $$log_{3}(x^{2} + 2x)$$ And re-write the original equation: $$log_{3}(x^{2} + 2x) = 1$$ Finally, we can change the new logarithm into exponential form and solve accordingly: $$log_{3}(x^{2} + 2x) = 1$$ $$3^{1} = x^{2} + 2x$$ $3 = x^{2} + 2x$ $0 = x^{2} + 2x - 3$ which can be factorized since two factors of $3$ that, when substracted, yield $2$, are $3\times1$: $0 = (x + 3)(x - 1)$; where $x = -3$ and $x = 1$. Since $log_{3}(-3)$ is not a defined function, the only valid solution is $x = 1$.
