Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 24

log_3(x) + log_3(x - 2) = 1
The logarithms at the left side have the same base. So express the left side with one logarithm only using the rule log_b (M) + log_b (N) = log_b(M*N ).
log_3(x * (x-2)) = 1
log_3 (x^2 - 2x) = 1
Then, convert this to exponential form.
Take note that if a logarithmic equation is in the form
y = log_b (x)
its equivalent exponential equation is
x = b^y
So converting
log_3 (x^2 - 2x) =1
to exponential equation, it becomes:
x^2-2x = 3^1
x^2 - 2x = 3
Now the equation is in quadratic form. To solve it, one side should be zero.
x^2 - 2x - 3 = 0
Factor the left side.
(x - 3)(x +1)=0
Set each factor equal to zero. And isolate the x.
x - 3 = 0
x=3

x + 1=0
x = -1
Now that the values of x are known, consider the condition in a logarithm. The argument of a logarithm should always be positive.
In the equation
log_3(x) + log_3(x - 2)=1
the arguments are x and x - 2. So the values of these two should all be above zero.
x gt 0
x - 2gt0
Between the two values of x that we got, it is only x = 3 that satisfy this condition.

Therefore, the solution is x=2 .