9. Solve for x and check:

log12x + log12(x + 1) = 1

First combine a a single log.

log12x(x + 1) = 1

Now we can change to exponential notation.

x(x + 1) = 121

Simplify. Remove parentheses on the left and compute the power on the right.

x2 + x = 12

This gives us a quadratic. Transpose everything to one side, leaving a 0 on the other and hope that it factors.

x2 + x - 12 = 0

It does

(x - 3)(x + 4) = 0

Set the factors = 0 and solve.

x - 3 = 0

x + 4 = 0

x = 3

x = -4

We must now check the solutions. If x = 3, the eauation becomes

log123 + log12(3 + 1) = 1

log123(3 + 1) = 1

log1212 = 1

1 = 1

and it checks. For the other solution, let x = -4

log12(-4) + log12(-4 + 1) = 1

log12(-4) + log12(-3) = 1

Here we are being asked to take logs of negative numbers, so the solution doesn't check.

The only solution, then, is

x = 3