Question

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

Answer 1

`x=(5+sqrt21)/2` and `x=(5-sqrt21)/2`

Explanation:

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

`log(a)+log(b)=log(ab)`

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

`log(x(5-x))=0`

which can be simplified to

`log(-x^2+5x)=0`

We essentially now have

`log_10(-x^2+5x)=log_10color(blue)((10^0))`

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(-x^2+5x)=log_10(1)`

Both sides of the equation have `log_10`, so they cancel.

`cancel(log_10)(-x^2+5x)=cancel(log_10)(1)`

What's left is

`-x^2+5x=1`

Turning this into a quadratic, we get

`-x^2+5x-1=0`

We can now apply the Quadratic Formula, `(-b+-sqrt(b^2-4ac))/(2a)`

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

Plugging in, we get

`x=(-5+-sqrt(25-4))/-2`

Which simplifies to

`(5+-sqrt21)/2`

Thus, our roots are equal to

`x=(5+sqrt21)/2` and `x=(5-sqrt21)/2`

Hope this helps!