For what values of x is the expression log(x^2 - 4x- 5 + 2x)/sqrt(x - 1) defined? Express your answer in interval notation.
To determine the values of x for which the given expression is defined, we need to consider two aspects:
The argument of the logarithm (logarithm domain).
The denominator involving the square root.
Let's address each aspect separately:
The argument of a logarithm must be greater than zero for the logarithm to be defined. In this case, we have the argument as:
x^2 - 4x - 5 + 2x
We need to find when this expression is greater than zero:
x^2 - 4x - 5 + 2x > 0
Simplify the inequality:
x^2 - 2x - 5 > 0
Now, we can find the values of x that satisfy this inequality. You can use methods like factoring or the quadratic formula, but I'll provide the factorized form:
(x - 3)(x + 1) > 0
To determine the sign of the expression on the left-hand side, we can create a sign chart by considering the signs of each factor:
(x - 3) is positive for x > 3 and negative for x < 3.
(x + 1) is positive for x > -1 and negative for x < -1.
Now, we need the product of these factors to be greater than zero, which means both factors must have the same sign. This occurs when:
x > 3 and x > -1 (since both factors are positive).
Therefore, for the logarithm to be defined, x must be greater than 3.
Square Root Denominator:
The denominator of the square root is:
sqrt(x - 1)
For the denominator to be defined, the expression inside the square root (x - 1) must be greater than or equal to zero:
x - 1 >= 0
Solve for x:
x >= 1
Now, we have two conditions:
For the logarithm to be defined: x > 3.
For the denominator (square root) to be defined: x >= 1.
To find the values of x for which the entire expression is defined, we consider the intersection of these two conditions:
x > 3 (logarithm condition) and x >= 1 (denominator condition).
The values of x that satisfy both conditions are x >= 3. Therefore, the expression is defined for all x in the interval [3, ∞).