#log_2(x-3) = 2- log_2(x-6)# What is x?

#log_2(x-3) = 2 - log_2(x-6)#
#log_2(x-3) + log_2(x-6) = 2#
#log_2((x-3)(x-6)) = 2#
#log_2(x^2-6x-3x+18) = 2#
#log_2(x^2-9x+18) = 2#
#x^2 - 9x + 18 = 4#
#x^2 -9x +14 = 0#

Now that the equation is in the form #ax^2 + bx + c = 0#, the quadratic formula (#x = (-b+-sqrt(b^2-4ac))/(2a)#) can be applied:
#x = (-(-9)+-sqrt((-9)^2-4(1)(14)))/(2(1))#
#x=(9+-sqrt(81-56))/2#
#x=(9+-sqrt(25))/2#
#x = (9+-5)/2#

Adding #5# for the first root of #x#:
#x = (9+5)/2 = 14/2 = 7#

Subtracting #5# for the second root of #x#:
#x = (9-5)/2 = 4/2 =2#

However, a #log# can not be applied to a negative number, so there are two more conditions for the root(s) of x:
#(x-3)>0# and #(x-6)>0#

Testing #7# to see if it is a root of #x#;
#(7-3)>0#, #4>0# The first condition is met.
#(7-6)>0#, #1>0# the second condition is met.
Thus, #7# is a root of #x#.

Testing #2# to see if it is a root of #x#:
#(2-3)>0# Since #-1# is not more than #0#, #2# is not a root of #x#.

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logy=log(xy)#

#•color(white)(x)log_b x=nhArrx=b^n#

#"add "log_2(x-6)" to both sides"#

#rArrlog_2(x-3)+log_2(x-6)=2#

#rArrlog_2(x-3)(x-6)=2#

#rArr(x-3)(x-6)=2^2=4#

#rArrx^2-9x+14=0larrcolor(blue)"in standard form"#

#"the factors of + 14 which sum to - 9 are - 2 and - 7"#

#rArr(x-2)(x-7)=0#

#"equate each factor to zero and solve for x"#

#x-2=0rArrx=2#

#x-7=0rArrx=7#

#(x-3)>0" and "(x-6)>0#

#rArrx=2" is invalid"#

#rArrx=7" is the solution"#