Simplify by writing as a single logarithm & State restrictions:
log(6𝑥^2+5𝑥−6) + log(2x − 3) − log(4𝑥^2 − 9)
since log(ab) = loga - logb, you have
log (2x+3)(3x-2) + log (2x+3) - log (2x-3)(2x+3)
= log(2x+3) + log(3x-2) + log(2x+3) - log(2x-3) - log(2x+3)
= log (2x+3)(3x-2)/(2x-3)
Restrictions: Using the original problem's statement, each quantity must be positive, since logu is defined only if u is positive.
So, using the first term, you need
6x^2+5x-6 > 0
(2x+3)(3x-2) > 0
x must lie in (-∞,-3/2)U(2/3,∞)
Now find the other two intervals and use their intersection, since all three must be positive.
To simplify the expression log(6𝑥^2+5𝑥−6) + log(2x − 3) − log(4𝑥^2 − 9) as a single logarithm, we can use the logarithmic properties.
The first property we will use is the product rule of logarithms, which states that log(a) + log(b) = log(a * b).
Applying this rule to the given expression, we have:
log(6𝑥^2+5𝑥−6) + log(2x − 3) − log(4𝑥^2 − 9)
= log((6𝑥^2+5𝑥−6) * (2x − 3)) - log(4𝑥^2 − 9)
To simplify further, we can use another property called the quotient rule of logarithms, which states that log(a) - log(b) = log(a / b).
Using this property, we have:
log((6𝑥^2+5𝑥−6) * (2x − 3)) - log(4𝑥^2 − 9)
= log((6𝑥^2+5𝑥−6) * (2x − 3) / (4𝑥^2 − 9))
Now, to write this expression as a single logarithm, we can combine the numerator and denominator using parentheses:
= log(((6𝑥^2+5𝑥−6) * (2x − 3)) / (4𝑥^2 − 9))
This is the simplified expression as a single logarithm.
Now, let's discuss the restrictions. In logarithmic expressions, we need to ensure that the argument of the logarithm is positive and not equal to zero. In this case, we have three arguments: (6𝑥^2+5𝑥−6), (2x − 3), and (4𝑥^2 − 9).
To find the restrictions, we need to consider each argument individually:
1. For (6𝑥^2+5𝑥−6):
We need to ensure that 6𝑥^2+5𝑥−6 > 0.
We can solve this quadratic inequality by factoring or using the quadratic formula.
The factored form of the quadratic is (2𝑥−1)(3𝑥+6), so we have two critical points: x = 1/2 and x = -6/3 = -2.
By testing each interval separately, we find that the quadratic is positive when x < -2 and x > 1/2.
2. For (2𝑥 − 3):
We need to ensure that 2𝑥 − 3 > 0.
Solving this inequality, we find that the critical point is x = 3/2, and since the coefficient of x is positive (2), the inequality is satisfied when x > 3/2.
3. For (4𝑥^2 − 9):
We need to ensure that 4𝑥^2 − 9 > 0.
This is a difference of squares, which factors as (2𝑥 − 3)(2𝑥 + 3). Thus we have two critical points: x = 3/2 and x = -3/2.
By testing each interval separately, we find that the inequality is satisfied when x < -3/2 or x > 3/2.
Therefore, the restrictions for the given expression are:
x < -2, x > 1/2, and x < -3/2 or x > 3/2.