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:


I'll first show you a way to solve the multiple choice question, then I'll explain the general answer.
Do not forget to check the initial conditions! The function $\log x$ exists, over the reals, if and only if $x > 0$. Therefore you have that the final solution $x$ must satisfy $x > 2$. With this you can throw away most of the answers. The only one remaining is $3$, which, under normal conditions, would be the correct one. However, just by plugging it in the equation we see that it cannot be the answer.
I think that either there is an error in the book or that you didn't read it correctly.
In general, to solve a quadratic equation you can always make use of the quadratic formula. You first have to transform you equation to the canonical form
$$ax^2 + bx + c = 0.$$
In your case, we have $x^2 - 3x - 98 = 0$, and the quadratic formula tells us that
$$x = \frac32 \pm \frac{\sqrt{401}}2.$$
You have to discard the solution with the minus because, as you will remember, there are some initial conditions for the existence of the logarithm.
