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

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Answer 1 · 2016-01-11T11:21:29

#x=50#

#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)#