SOLUTION: find the domain of the logarithmic function log 3 (2x+2)+log 3 (x+3)=log 3 x

Question 709749: find the domain of the logarithmic function log 3 (2x+2)+log 3 (x+3)=log 3 x
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Sometimes the domain of a function is given to you. But when you are asked to find the (implicit) domain then you:

Start by assuming that the domain is all real numbers.
Look for x values that "cannot be". If you find some then exclude these values from the domain.

What kinds of x values "cannot be"? What kinds of x values must be excluded from the domain? Answer: There are a variety of "no-no's" in Math:

Zero denominators. Any x that would make a denominator in the function turn into zero must be excluded.
Negative radicands of even-numbered roots. ("Radicand" is the name for the expression inside a radical. "Even-numbered roots" means square (or 2nd) roots, 4th roots, 6th roots, etc.) Since you cannot get a negative result when raising a real number to an even power, we cannot have negative radicands in even-numbered roots. Any x values that would make the radicand of an even-numbered root turn negative must be excluded from the domain.
Invalid arguments or bases of logarithms. Valid arguments to logarithms are positive numbers. Valid bases for logarithms are positive but not a 1. And x value that makes the argument or base of a log be invalid must be excluded from the domain.
Certain arguments to certain Trig functions and inverse Trig functions are invalid/undefined. And x value that would make the argument of any Trig function (or inverse) invalid/undefined must be excluded from the domain.

These are the most "popular" reasons for excluding x values from the domain. In a nutshell: Any x value(s) that make some part of a function's definition invalid/undefined must be excluded from the domain of the function.

Your function has no denominators, even-numbered roots or Trig functions. But it does have logarithms. The bases have no x's in them so they will be valid no matter what x is. But the arguments have x's in them so we must make sure they are valid. We need all three arguments to be positive. So:
2x + 2 > 0 and x + 3 > 0 and x > 0
Solving these we get:
x > -1 and x > -3 and x > 0
These all say x is greater than something. In this situation we can replace all three with just one: x > 0. (Take a moment to see why this is so. If x > 0, won't it automatically be greater than -1 and -3? On the other hand, if x > -1 then x will be greater than -3 but it won't necessarily be greater than zero. (e.g. -1/2 is greater than -1 but not greater than zero.))

So the domain of f(x) is all numbers greater than zero (IOW: all positive numbers).
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