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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