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

1 Answer
Jun 2, 2018

#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!