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

Mar 3, 2018

The solution is $x = 2$.

#### Explanation:

${\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{red}{x}\right) + {\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{b l u e}{y}\right) = {\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{red}{x} \textcolor{b l u e}{y}\right)$

Now combine the two $\log$s, then rewrite $1$ as $\log \left(10\right)$, then cancel the logs on either side:

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

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

$\log \left({x}^{2} + 3 x\right) = 1$

$\log \left({x}^{2} + 3 x\right) = \log \left(10\right)$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\log}}} \left({x}^{2} + 3 x\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{\log}}} \left(10\right)$

${x}^{2} + 3 x = 10$

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

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

$x = 2 , - 5$

Plug in the answers to the original equation and see if they still work:

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

Testing $2$:

$\log \left(2\right) + \log \left(2 + 3\right) = 1$

$\log \left(2\right) + \log \left(5\right) = 1$

$0.30102 \ldots + 0.69898 \ldots = 1$

1=1qquadqquad color(lightgreen)sqrt

The solution $2$ works. Testing $- 5$:

$\log \left(- 5\right) + \log \left(- 5 + 3\right) = 1$

Since $\log \left(- 5\right)$ is undefined, this solution doesn't work. Therefore, the only solution is $x = 2$.