log(x-2) + log(9-x) < 1

log(x-2) + log(9-x) < 1
log(x-2)(9-x)<1
(x-2)(9-x)<10^1
and solve.

To solve the inequality log(x-2) + log(9-x) < 1, we will need to use logarithmic properties and algebraic manipulations. Here's how we can approach it step by step:

Step 1: Combine the logarithms using the product rule of logarithms.
log(x-2)(9-x) < 1

Step 2: Remove the logarithm by exponentiating both sides of the inequality.
10^log(x-2)(9-x) < 10^1
(x-2)(9-x) < 10

Step 3: Expand the expression on the left-hand side.
9x - x^2 - 18 + 2x < 10
-x^2 + 11x - 28 < 10

Step 4: Rearrange the inequality into standard quadratic form.
-x^2 + 11x - 28 - 10 < 0
-x^2 + 11x - 38 < 0

Step 5: Solve the quadratic equation by factoring or using the quadratic formula. In this case, we will use factoring.
(x - 2)(-x + 19) < 0

Step 6: Set up a sign chart by selecting test points to determine the signs of each factor.
Test x = 2: (2 - 2)(-2 + 19) < 0 --> 0 < 0 (FALSE)
Test x = 19: (19 - 2)(-19 + 19) < 0 --> 17 < 0 (TRUE)

Step 7: From the sign chart, identify the interval(s) where the expression is less than zero (negative).
From the sign chart, we can see that the inequality is satisfied when x is between 2 and 19.

Therefore, the solution to the inequality log(x-2) + log(9-x) < 1 is 2 < x < 19 (interval notation) or (2, 19) (set builder notation).