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

Jan 24, 2016

I found $x = 5$ if the base of the log is $10$.

#### Explanation:

We can take advantage of the property of logs that says:
$\log \left(x\right) + \log \left(y\right) = \log \left(x y\right)$
to get in your case:
$\log \left[x \left(x - 3\right)\right] = 1$
Now the problem is the base of your log...
$\textcolor{red}{\to}$ If the base is $10$ we write:
${\log}_{10} \left[x \left(x - 3\right)\right] = 1$
apply the definition of log:
$x \left(x - 3\right) = {10}^{1}$
solve for $x$:
${x}^{2} - 3 x - 10 = 0$
apply the Quadratic Formula:
${x}_{1 , 2} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2}$
two solutions:
${x}_{1} = \frac{3 + 7}{2} = 5$
${x}_{2} = \frac{3 - 7}{2} = - 2$ NOT because you'd get a negative argument of the log in the original expression.

If the base is NOT $10$ use your specific base at the above stage marked:
$\textcolor{red}{\to}$