#log(x-9)=3-log(100x)#
By the laws of logarithms:
#log(ab)=log(a)+log(b)color(white)(888)[1]#
#log(100x)=log(100)+log(x)#
Assuming these are base 10 logarithms:
#log(100)+log(x)=2+log(x)#
We now have:
#log(x-9)=3-2-log(x)#
#log(x-9)=1-log(x)#
Using #[1]#
#log(x-9)+log(x)=1#
#log(x(x-9))=1#
#log(x^2-9x)=1#
#10^(log(x^2-9x))=10^(1)#
#x^2-9x=10#
#x^2-9x-10=0#
Factor:
#(x+1)(x-10)=0=>x=-1 and x=10#
Checking solutions.
#x=-1#
#log((-1)-9)=3-log(100(-1))#
#log(-10)=3-log(-100)#
Logarithms are only defined for real numbers if for:
#log(x)#
#x>0#
Therefore #-1# is not a solution.
For #x=10#
#log(10-9)=3-log(100(10))#
#log(1)=3-log(1000)#
#0=3-3#
#0=0#
So #x=10# is the only solution.