Solving Logarithmic Equations Finding Integral Values Of A

Hey there, math enthusiasts! Ever stumbled upon a logarithmic equation that seems like a tricky puzzle? Well, today, we're diving deep into one such equation and cracking it open. We're going to explore the equation ln(x^2 + 5x) - ln(x + a + 3) = 0 and figure out the possible integral values of 'a' that make this equation have exactly one solution for 'x'. Buckle up, because this is going to be an exciting mathematical journey!

The Challenge: Solving the Logarithmic Equation

So, guys, let's break down this challenge. We're given the equation ln(x^2 + 5x) - ln(x + a + 3) = 0. Our mission, should we choose to accept it (and we totally do!), is to find the values of 'a' that make this equation have only one solution. Now, logarithmic equations can be a bit finicky, so we need to be careful about the domain and the properties of logarithms.

Understanding the Logarithmic Domain

First things first, let's talk about the domain. Remember, the logarithm of a non-positive number is undefined. This means that the arguments of our logarithms, x^2 + 5x and x + a + 3, must be strictly greater than zero. This is a crucial starting point because it limits the possible values of 'x' that can be solutions to our equation.

The domain of ln(x^2 + 5x) is where x^2 + 5x > 0. Factoring this quadratic, we get x(x + 5) > 0. This inequality holds when both factors are positive or both are negative. So, we have two cases:

1. x > 0 and x + 5 > 0 (which means x > 0)
2. x < 0 and x + 5 < 0 (which means x < -5)

Therefore, the domain for the first logarithm is x ∈ (-∞, -5) ∪ (0, ∞). This means x must be less than -5 or greater than 0.

Next, let's consider the domain of ln(x + a + 3). We need x + a + 3 > 0, which implies x > -a - 3. This gives us another constraint on the possible values of 'x', and it depends on the value of 'a'.

Simplifying the Equation

Now that we've tackled the domain, let's simplify the equation itself. We can use the logarithmic property ln(A) - ln(B) = ln(A/B) to combine the two logarithms:

ln(x^2 + 5x) - ln(x + a + 3) = ln((x^2 + 5x) / (x + a + 3)) = 0

To get rid of the logarithm, we can exponentiate both sides with base 'e' (since we're dealing with natural logarithms):

e^(ln((x^2 + 5x) / (x + a + 3))) = e^0

This simplifies to:

(x^2 + 5x) / (x + a + 3) = 1

Now we have a rational equation that we can work with. Multiplying both sides by (x + a + 3) gives us:

x^2 + 5x = x + a + 3

Rearranging the terms, we get a quadratic equation:

x^2 + 4x - (a + 3) = 0

Cracking the Quadratic: Finding the Conditions for One Solution

Okay, we've transformed our logarithmic equation into a quadratic equation. Now, we need to figure out the conditions for this quadratic to have exactly one solution. Remember, a quadratic equation has the general form Ax^2 + Bx + C = 0, and its solutions are given by the quadratic formula:

x = (-B ± √(B^2 - 4AC)) / (2A)

The discriminant, Δ = B^2 - 4AC, plays a crucial role here. If:

• Δ > 0: The quadratic has two distinct real solutions.
• Δ = 0: The quadratic has exactly one real solution (a repeated root).
• Δ < 0: The quadratic has no real solutions.

In our case, A = 1, B = 4, and C = -(a + 3). So, the discriminant is:

Δ = 4^2 - 4(1)(-(a + 3)) = 16 + 4(a + 3) = 16 + 4a + 12 = 4a + 28

For the quadratic to have exactly one solution, we need Δ = 0:

4a + 28 = 0

Solving for 'a', we get:

4a = -28

a = -7

So, one possible value for 'a' is -7. But hold on, we're not done yet! We need to make sure this solution is valid within the domain of our original logarithmic equation.

Checking the Domain: Ensuring a Valid Solution

We found that a = -7 makes the discriminant zero, giving us one solution for the quadratic. But we need to check if this solution also satisfies the domain conditions we established earlier.

With a = -7, our quadratic equation becomes:

x^2 + 4x - (-7 + 3) = x^2 + 4x + 4 = 0

This factors nicely as:

(x + 2)^2 = 0

So, the single solution for 'x' is x = -2. Now, let's check if this value is within the domain of our original logarithmic equation.

Remember, the domain of ln(x^2 + 5x) is x ∈ (-∞, -5) ∪ (0, ∞). Since -2 is not in this domain, it's not a valid solution.

This is a crucial point, guys! We found a value of 'a' that gives us one solution for the quadratic, but that solution doesn't work in the original logarithmic equation because it's outside the domain. This means we need to consider another scenario.

Another Scenario: When One Root is Extraneous

What if the quadratic has two solutions, but one of them is extraneous (i.e., it doesn't satisfy the domain conditions)? In this case, we would still have only one valid solution for the original logarithmic equation.

For the quadratic to have two distinct real solutions, we need Δ > 0:

4a + 28 > 0

4a > -28

a > -7

So, if a > -7, the quadratic has two solutions. Let's call them x1 and x2. We need one of these solutions to be outside the domain x ∈ (-∞, -5) ∪ (0, ∞) and the other one to be inside.

Let's think about the other domain condition: x > -a - 3. If one of the solutions is less than or equal to -5 or between (-5,0], it will be extraneous.

To find the solutions x1 and x2, we use the quadratic formula:

x = (-4 ± √(4a + 28)) / 2 = -2 ± √(a + 7)

So, x1 = -2 + √(a + 7) and x2 = -2 - √(a + 7). Notice that x2 will always be less than x1.

Let's analyze the conditions for one solution to be extraneous. We need to consider two cases:

1. Case 1: x2 ≤ -5

   -2 - √(a + 7) ≤ -5

   -√(a + 7) ≤ -3

   √(a + 7) ≥ 3

   a + 7 ≥ 9

   a ≥ 2

   If a ≥ 2, then x2 ≤ -5. We also need to make sure that x1 is within the domain, i.e., x1 > 0:

   -2 + √(a + 7) > 0

   √(a + 7) > 2

   a + 7 > 4

   a > -3

   Since a ≥ 2, this condition is satisfied.

2. Case 2: x1 ≤ 0 If x1 is less than or equal to 0, the root is extraneous -2 + √(a + 7) ≤ 0

   √(a + 7) ≤ 2

   a + 7 ≤ 4

   a ≤ -3

If x1<=0, then x1 cannot be greater than 0 and less than -5. So this range is not possible

The Integral Values of 'a': Our Final Answer

Alright, guys, we've done some serious math detective work here! Let's summarize our findings:

• We found that a = -7 makes the discriminant zero, but the resulting solution for 'x' was extraneous.
• We then considered the case where the quadratic has two solutions, but one of them is extraneous.
• We found that if a ≥ 2, one solution (x2) is less than or equal to -5 and the other solution (x1) will be greater than 0

Therefore, the possible integral values of 'a' that give the equation exactly one solution are all integers greater than or equal to 2. For example, a can be 2, 3, 4, and so on.

Conclusion: Math is an Adventure!

So, there you have it! We've successfully navigated the world of logarithmic equations, quadratic equations, domains, and extraneous solutions. This problem really highlights the importance of not just solving equations, but also understanding the underlying concepts and checking the validity of our solutions. Math can be a wild ride, but it's definitely an adventure worth taking!