
William W. answered 11/07/19
Top Algebra Tutor
This is a little strange in that we don't typically ask for the domain of an equation. But the inference is that both sides of the equal sign are functions, each with its own domain and you are trying to find the resulting domain that is most restrictive for both functions. So that's what we'll do:
Initial explanation: When we write logb(m) = n what we are saying is that b raised to the n power is m or (writing it mathematically) bn = m. Notice that there is no exponent "n" that you can pick that will result in "m" being a negative number, or even zero. So "m" MUST ALWAYS BE > 0.
For the first problem, given the information above, we can say that the x values we pick MUST make both "x2 - 2x" > 0 and also must make "2x + 12" > 0 so let's solve those to see what x must be.
First:
x2 - 2x > 0
x(x - 2) >0 This is true when both "x" and "x - 2" are greater than zero or when both "x" and "x - 2" are less than zero. They are both greater than zero when x > 2 and they are both less than zero when x < 0. So the simplified answer for this domain is -∞ < x < 0 or 2 < x < ∞
Second:
2x + 12 > 0
2x > -12
x > -6
This restricts the domain we found in the first part even more.
So, combining the two parts, the domain for the first problem is: -6 < x < 0 or 2 < x < ∞
For the second problem, "√x - 2" must be greater than zero
√x - 2 > 0
√x > 2
x > 4
The second function (right hand side of the equal sign) is just a constant function, it doesn't matter what you pick for the input, the output is always 1 so this doesn't restrict the domain we found for the left side of the equal sign. So the domain for the second problem is x > 4