Question 588203
log(x-6)- log (x-2) = log 5/x
log(x-6)- log (x-2) - log 5/x=0
log(x-6)- (log (x-2) + log 5/x)=0
place under single log
log[(x-6)/(x-2)*5/x]=0
convert to exponential form: base(10) raised to log of number(0)=number[(x-6)/(x-2)*5/x]
10^0=[(x-6)/(x-2)*5/x]=1
x-6=(x-2)*5/x=(5x-10)/x
x^2-6x=5x-10
x^2-11x+10=0
(x-10)(x-1)=0
x=1 (reject, (x-2)>0)
or 
x=10