How do you solve Log x + log(x+3)= log 10?

Jul 19, 2015

Use properties of $\log$, exponentiation and solve a quadratic to find solution $x = 2$.

Explanation:

$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

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

Note we require $x > 0$ in order that $\log x$ is defined.

So:

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

Take exponent base $10$ of both sides to get:

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

Subtract $10$ from both sides to get:

$0 = {x}^{2} + 3 x - 10 = \left(x + 5\right) \left(x - 2\right)$

So $x = - 5$ or $x = 2$.

$x = - 5$ is spurious, because we require $x > 0$ in order that $\log x$ be defined.

So the only valid solution is $x = 2$