since log(a) + log(b) = log(a*b), this equation becomes:
log((x-2)*(9-x)) < 1
simplify to get log(9x - x^2 -18 + 2x) < 1
combine like terms and reorder the expression by descending order of degree to get log(-x^2 + 11x - 18) < 1
since log(a) = b if and only if 10^b = a, then you get:
this is true if and only if -x^2 + 1x - 18 < 10^1 which can also be shown as:
this is true if and only if 10^1 > -x^2 + 11x - 18.
since 10^1 = 10, this equation becomes 10 > -x^2 + 11x - 18.
add x^2 + 18 to both sides of the equation and subtract 11x from both sides of the equation and you get x^2 - 11x + 18 + 10 > 0.
combine like terms to get x^2 - 11x + 28 > 0
set the equation equal to 0 and factor the quadratic to get (x-4) * (x-7) = 0
solve for x to get x = 4 and x = 7.
you know the equation is equal to 0 when x = 4 and x = 7.
solve for the different regions created to determine when the equation is greater than 0 and when the equation is less than zero.
you will find that the equation is greater than 0 when x < 4 and when x > 7.
this is the graph of that equation.
you could be satisfied and say that the original equation is satisfied with x < 4 and when x > 7, except that you are dealing with logs, and the expression within the log has to be > 0.
the original equation is log(x-2) + log(9-x) < 1
in order for the expression inside the log to be greater than 0, then x has to be greater than 2 and x has to be smaller than 9.
now you have 4 requirements.
they are:
x > 2
x < 9
x < 4
x > 7
this puts x in the following ranges.
x > 2 and x < 4
x > 7 and x < 9
that would be your answer.
if you graph y = the original equation and you graph y = 1, you should see that the original equation is satisfied when x > 2 and < 4, and when x > 7 and < 9.
that graph is shown below:
you can see from the graph that the log equation is less than 1 when x is greater than 2 and less than 4, and when x is greater than 7 and less than 9.
