How do you solve #log(x-3)+log x=1#?

#log(x−3)+logx=1#

logarithms on the left side may be added together by multiplying argumnts, on the other side we can rewrite number 1

#log[(x−3)*x]=log10#

Logarithm is a simple function, therefore we can compare arguments

#[(x−3)*x]=10#

#x^2−3x-10=0#

#(x+2)*(x-5)=0#

#x_1=-2#
#x_2=5#

Since argument of logaritm can be only positive, #x_1# is not a solution
#log(-2−3)+log(-2)=># not possible

#log(5−3)+log5=>log2+log5=>log(5*2)=>log10=>1#
#1=1# ;correct;

The goal with type of problems is to get a expresion like #logA=logB#. By injectivity of function log, we can say that #A=B#

Let see...
Using logarithmic rules

#log(x-3)+logx=log10=1#

#log(x-3)x=log10#

Then #(x-3)x=10#

However 10=5x2 or 2x5, or negatives ones. Let say #x-3=5# then #x=2#

If #x=5# then #x-3=2#

Other way is operate in #x(x-3)=x^2-3x=10# or

#x^2-3x-10=0# using quadratic formula we have

#x=(3+-sqrt(9+40))/2=(3+-7)/2# this arrives to #x=5# and #x=-2#

Lets check the answers #log(-2-3)+log-2=1# reject this solution because nor #log-5# neither #log-2# doesn`t exists

By other hand #log(5-3)+log5=log(2xx5)=log10=1#

So, the only solution is #x=5#