Solving Logarithms: Finding The Domain

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Solving Logarithms: Finding the Domain

Hey guys! Let's dive into the fascinating world of logarithms and figure out how to determine the values of x for which these logarithmic expressions are actually defined. This is super important because logarithms, like any mathematical function, have specific rules about what's allowed. We can't just plug in any old number and expect a valid answer. So, we're going to break down several examples, making sure we understand the conditions that must be met for a logarithm to exist. Get ready to flex those math muscles!

a) log_(x+2) (15+2x-x^2)

Alright, let's start with our first example: log_(x+2) (15+2x-x^2). Here, we need to consider two key aspects. First, the base of the logarithm, which is x+2, must be positive and not equal to 1. This is the golden rule of logarithms! Second, the argument of the logarithm, which is 15+2x-x^2, must also be positive. Let's tackle these one by one, step-by-step:

  1. Base Condition: x + 2 > 0 and x + 2 ≠ 1. Solving x + 2 > 0, we get x > -2. For x + 2 ≠ 1, subtracting 2 from both sides gives us x ≠ -1. So, for the base, we have x > -2 and x ≠ -1.
  2. Argument Condition: 15 + 2x - x^2 > 0. To solve this, let's first rearrange the inequality: -x^2 + 2x + 15 > 0. It's often easier to work with a positive leading coefficient, so we multiply both sides by -1. Remember, this flips the inequality sign! We get x^2 - 2x - 15 < 0. Now, let's factor the quadratic expression: (x - 5)(x + 3) < 0. To find the solution, let's think about the parabola represented by y = (x - 5)(x + 3). It opens upwards, and it crosses the x-axis at x = 5 and x = -3. The expression is negative (less than zero) between these two roots. Therefore, the solution for the argument condition is -3 < x < 5.
  3. Combining Conditions: Now, we need to find the intersection of the base and argument conditions. We have x > -2, x ≠ -1, and -3 < x < 5. Combining these, we get the final solution: -2 < x < 5 and x ≠ -1. This means x can be any number between -2 and 5, but it cannot be -1. This process is key when solving problems, and this is why we must always follow the steps.

Now, let's move on to the next example and strengthen our skills!

b) log_(4-x)(x^2-x-6)

Okay, let's crank it up a notch with log_(4-x)(x^2-x-6). Remember our two conditions? They still apply. The base (4-x) must be positive and not equal to 1, and the argument (x^2-x-6) must be positive.

  1. Base Condition: 4 - x > 0 and 4 - x ≠ 1. Solving 4 - x > 0, we get x < 4. Solving 4 - x ≠ 1, we get x ≠ 3. So, for the base, we have x < 4 and x ≠ 3.
  2. Argument Condition: x^2 - x - 6 > 0. Let's factor this quadratic expression: (x - 3)(x + 2) > 0. Consider the parabola y = (x - 3)(x + 2). It opens upwards and crosses the x-axis at x = 3 and x = -2. The expression is positive (greater than zero) outside of these two roots. Therefore, the solution for the argument condition is x < -2 or x > 3.
  3. Combining Conditions: We need to find the intersection of x < 4, x ≠ 3, and (x < -2 or x > 3). Let's think about this on a number line. The solutions that satisfy both conditions are x < -2 or 3 < x < 4. This is the solution set for which the logarithm is defined. Always remember to consider the intersection.

Great job! You're really getting the hang of this. Let's press on and solve more examples!

c) log_(4x^2+5x+1) (x^2 +4)

Alright, let's tackle log_(4x^2+5x+1) (x^2 +4). Don't worry, even though the base looks a bit more complex, the rules remain the same! The base (4x^2+5x+1) must be positive and not equal to 1, and the argument (x^2 +4) must be positive.

  1. Base Condition: 4x^2 + 5x + 1 > 0 and 4x^2 + 5x + 1 ≠ 1. First, let's factor the quadratic: (4x + 1)(x + 1) > 0. Consider the parabola y = (4x + 1)(x + 1). It opens upwards and crosses the x-axis at x = -1 and x = -1/4. The expression is positive outside these roots, thus x < -1 or x > -1/4. Now, let's deal with the second part of the base condition: 4x^2 + 5x + 1 ≠ 1. This simplifies to 4x^2 + 5x ≠ 0. Factoring out an x, we get x(4x + 5) ≠ 0. This means x ≠ 0 and x ≠ -5/4. Therefore, for the base condition, we have x < -1 or -1/4 < x and x ≠ 0 and x ≠ -5/4.
  2. Argument Condition: x^2 + 4 > 0. This is always true because x^2 is always non-negative (greater than or equal to zero), and adding 4 makes it strictly positive. So, this condition doesn't restrict x.
  3. Combining Conditions: We need the intersection of x < -1 or x > -1/4, and x ≠ 0 and x ≠ -5/4. This gives us the final solution: x < -5/4 or -5/4 < x < -1 or -1/4 < x and x ≠ 0. We are always trying to find the values that make both statements true. Keep it up, you got it!

d) log_6 (sqrt(x+4)-3)

Let's get into log_6 (sqrt(x+4)-3). Here, the base is a constant (6), which is already positive and not equal to 1. So, we only need to worry about the argument, which is sqrt(x+4)-3.

  1. Argument Condition: sqrt(x+4) - 3 > 0. Let's solve this. Add 3 to both sides: sqrt(x+4) > 3. Now, square both sides: x + 4 > 9. Subtract 4 from both sides: x > 5. Additionally, the expression inside the square root must be non-negative: x + 4 >= 0, which means x >= -4. So, from the square root, it follows that x >= -4, then we combine this with the solution to the argument condition to satisfy the square root itself, and we get x > 5.
  2. Combining Conditions: There is only one condition, therefore the solution set is x > 5. The other condition x+4>=0 is included, and this is why the solution is x>5.

e) log_x sqrt(x^2-11x+18)

Okay, let's analyze log_x sqrt(x^2-11x+18). Remember, the base x must be positive and not equal to 1, and the argument sqrt(x^2-11x+18) must be positive.

  1. Base Condition: x > 0 and x ≠ 1.
  2. Argument Condition: sqrt(x^2 - 11x + 18) > 0. This means that x^2 - 11x + 18 > 0. Let's factor this quadratic: (x - 9)(x - 2) > 0. Consider the parabola y = (x - 9)(x - 2). It opens upwards and crosses the x-axis at x = 2 and x = 9. The expression is positive outside these roots. So, x < 2 or x > 9. We also need to consider the domain of the square root, meaning that x^2 - 11x + 18 >= 0. This is the same as the argument condition, except that it also allows for equality. Thus the solution for this is x <= 2 or x >= 9.
  3. Combining Conditions: We need to find the intersection of x > 0, x ≠ 1, and (x < 2 or x > 9). The solution is 0 < x < 2 and x ≠ 1, or x > 9.

f) log_[x] (2x-1)

Now, let's explore log_[x] (2x-1). Here, the base is the greatest integer function of x, denoted by [x]. Let's break this down carefully.

  1. Base Condition: [x] > 0 and [x] ≠ 1. If [x] > 0, then x >= 1. If [x] ≠ 1, this means x cannot be between 1 and 2 (excluding 2). So, x < 1 or x >= 2.
  2. Argument Condition: 2x - 1 > 0. Adding 1 to both sides and dividing by 2, we get x > 1/2.
  3. Combining Conditions: We need to find the intersection of the base and argument conditions. The solution is x > 1/2, x < 1 or x >= 2. Since x >= 2 implies that the greatest integer is greater than 0, then the final answer will be 1/2 < x < 1 or x >= 2. You are almost there, keep pushing!

g) lg [x+3]

Let's get into lg [x+3]. Here, lg denotes the base-10 logarithm. The base is 10, so it's fine. We only have to worry about the argument, which is [x+3].

  1. Argument Condition: [x+3] > 0. This means x + 3 >= 1, which means x >= -2.
  2. Combining Conditions: The final solution is x >= -2, because only this is needed to ensure the existence of the logarithm.

h) ln (3x+1-4)

Now, let's handle ln (3x+1-4). Here, ln denotes the natural logarithm (base e). The base is e, so it's good. Let's concentrate on the argument which is 3x+1-4.

  1. Argument Condition: 3x + 1 - 4 > 0. Simplifying, we have 3x - 3 > 0. Adding 3 to both sides: 3x > 3. Dividing by 3: x > 1.
  2. Combining Conditions: The solution is simply x > 1. Excellent!

i) lgDiscussion

It appears there's a typo, and this isn't a proper logarithmic expression. I will not answer. But you can use the previous rules to solve it. Great job, keep up the amazing work!