To solve the equation log7(x-1) + log7(x-2) = log7(x+2), we can combine the two logarithms on the left side using the product rule of logarithms, which states that log(a) + log(b) = log(ab).

So, log7((x-1)(x-2)) = log7(x+2)

Now, we can drop the logarithms since they have the same base and the equation becomes:

(x-1)(x-2) = x+2

Expanding the left side:

x^2 - 3x + 2 = x + 2

Rearranging the terms:

x^2 - 4x = 0

Now, we can factor out an x:

x(x - 4) = 0

So, the solutions are x = 0 and x = 4. However, we need to check if these solutions are valid by plugging them back into the original equation:

When x = 0:

log7(0-1) + log7(0-2) = log7(0+2)
This simplifies to:
log7(-1) + log7(-2) = log7(2)
Since the logarithm of a negative number is undefined, x = 0 is not a valid solution.

When x = 4:

log7(4-1) + log7(4-2) = log7(4+2)
This simplifies to:
log7(3) + log7(2) = log7(6)
This is a valid solution.

Therefore, the only valid solution to the equation log7(x-1) + log7(x-2) = log7(x+2) is x = 4.