How do you solve Log x + log (x-48) = 2?

1 Answer
Jan 11, 2016

x=50

Explanation:

logx+log(x-48)=2

The existence condition of log(g(x)) is:

g(x)>0

then

x>0

in system with:

x>48

:. x in(48,+oo)

using the logarithm product properties

log(x*(x-48))=2

Now we can resolve as follow:

10^(log(x*(x-48)))=10^2

x(x-48)=10^2
x^2-48x-100=0

x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(48+-sqrt(2304+400))/2=

=(48+-sqrt(2704))/2=(48+-52)/2=24+-26

x_1=-2

x_2=50

x_1 !in (48,+oo) is not a solution

x_2 in (48,+oo)