log(ab)=loga+logb
log3x+log[3(x-2)]=1
log[(3x)(3x-6)]=1
(3x)(3x-6)=10^1
9x^2-18x-10=0
find x and see if you would arrive at that
Solve the following algebraically.
log3(x) + log3(x-2) = 1
Are the solutions 3 and -1?
2 answers
clearly -1 is not a solution, since log(-1) is not real.
log3(x(x-2)) = 1
x^2-2x = 3
x^2-2x-3 = 0
(x-3)(x+1) = 0
so, yes, x=3 is a solution, but -1 is extraneous.
log3(x(x-2)) = 1
x^2-2x = 3
x^2-2x-3 = 0
(x-3)(x+1) = 0
so, yes, x=3 is a solution, but -1 is extraneous.
2 answers