How to solve $\log(x -1) + \log(x - 2) = 2?$ I'm doing this exercise:
$$\log(x - 1) + \log(x - 2) = 2$$
My steps:
Step 1: 
$$\log(x-1)(x - 2) = 2$$
Step 2:
$$(x - 1)(x - 2) = 10^2$$
Step 3:
$$x^2 - 3x + 2 = 100$$
Step 4:
$$x^2 = 98 + 3x$$
But I don't know what else I can do. In fact, I've doubts about the execution of $(x - 1)(x - 2)$?
Possible answers: 
a) 1 b) 0 c) 3
d) -2 e) -3
This is a print from the book:


You've done everything correctly so far, from $$x^2 = 3x +98 \iff x^2 - 3x - 98 = 0$$ which gives us (using the quadratic formula) $$x = \frac{3 \pm \sqrt{401}}{2}.$$
You'll have to discard the negative solution given the implicit impositions placed on $x$ through the logarithm. Now this doesn't match any of your options, which leads me to think that the original question is flawed in some way or there was a typo. 
