# How do you solve log x + log(5 – x) = 0?

Jun 2, 2018

$x = \frac{5 + \sqrt{21}}{2}$ and $x = \frac{5 - \sqrt{21}}{2}$

#### Explanation:

The key realization here is that we can use the logarithm property

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

Where $a = x$ and $b = 5 - x$. We can rewrite this logarithm as

$\log \left(x \left(5 - x\right)\right) = 0$

which can be simplified to

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

We essentially now have

${\log}_{10} \left(- {x}^{2} + 5 x\right) = {\log}_{10} \textcolor{b l u e}{\left({10}^{0}\right)}$

Notice, what we have in blue is also equal to $1$, so we didn't change the value of this equation. We now have

${\log}_{10} \left(- {x}^{2} + 5 x\right) = {\log}_{10} \left(1\right)$

Both sides of the equation have ${\log}_{10}$, so they cancel.

$\cancel{{\log}_{10}} \left(- {x}^{2} + 5 x\right) = \cancel{{\log}_{10}} \left(1\right)$

What's left is

$- {x}^{2} + 5 x = 1$

Turning this into a quadratic, we get

$- {x}^{2} + 5 x - 1 = 0$

We can now apply the Quadratic Formula, $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where $a = - 1 , b = 5 , c = - 1$

Plugging in, we get

$x = \frac{- 5 \pm \sqrt{25 - 4}}{-} 2$

Which simplifies to

$\frac{5 \pm \sqrt{21}}{2}$

Thus, our roots are equal to

$x = \frac{5 + \sqrt{21}}{2}$ and $x = \frac{5 - \sqrt{21}}{2}$

Hope this helps!