Finding Values Of X For Defined Logarithms A Comprehensive Guide

October 18, 2025 by Scholario Team 65 views

Hey guys! Today, we're diving deep into the fascinating world of logarithms and figuring out just when these mathematical expressions make sense. We're going to tackle a bunch of examples, so you'll be a pro at determining the values of x that keep our logs happy and defined. So, grab your thinking caps, and let's get started!

Understanding the Basics of Logarithms

Before we jump into the problems, let's quickly recap what logarithms are all about. A logarithm is basically the inverse operation to exponentiation. Think of it this way: if we have an equation like b^y = x, the logarithm answers the question, "What exponent (y) do I need to raise the base (b) to, in order to get x?" This is written as log*b(x) = y.

Now, here’s the crucial part: for a logarithm to be defined, we need to meet a few key conditions:

The base (b) must be positive and not equal to 1. This is because if the base were 1, 1 raised to any power is still 1, so we wouldn't get a unique exponent. If the base were negative or zero, we'd run into all sorts of issues with exponents, especially non-integer exponents.
The argument (x) must be positive. You can't take the logarithm of a negative number or zero because there's no exponent you can raise a positive base to and get a non-positive result.

With these rules in mind, let's dive into the examples and see how to apply them!

Solving for x in Various Logarithmic Expressions

a) log(x-2)3

Okay, first up, we've got log(x-2)3. Remember our rules? The base (x-2) needs to be positive and not equal to 1, and the argument (3) is already positive, so we're good there. Let's break it down:

Base must be positive: x - 2 > 0
- Adding 2 to both sides, we get x > 2.
Base cannot be 1: x - 2 ≠ 1
- Adding 2 to both sides, we get x ≠ 3.

So, for this logarithm to be defined, x must be greater than 2, but it can't be 3. We can write this in interval notation as x ∈ (2, 3) ∪ (3, ∞).

b) log(x-1)(5-x)

Next, we have log(x-1)(5-x). This one's a bit trickier because we have x in both the base and the argument. Let's tackle each condition:

Base must be positive: x - 1 > 0
- Adding 1 to both sides, we get x > 1.
Base cannot be 1: x - 1 ≠ 1
- Adding 1 to both sides, we get x ≠ 2.
Argument must be positive: 5 - x > 0
- Adding x to both sides, we get 5 > x, or x < 5.

Now, we need to combine these conditions. x must be greater than 1, not equal to 2, and less than 5. In interval notation, this is x ∈ (1, 2) ∪ (2, 5).

c) log(x-1)(16-x²)

Here we have log(x-1)(16-x²). Similar to the previous one, we have x in both the base and argument, but this time, the argument is a quadratic. Let's break it down:

Base must be positive: x - 1 > 0
- Adding 1 to both sides, we get x > 1.
Base cannot be 1: x - 1 ≠ 1
- Adding 1 to both sides, we get x ≠ 2.
Argument must be positive: 16 - x² > 0
- This is the same as x² < 16.
- Taking the square root of both sides (and remembering to consider both positive and negative roots), we get -4 < x < 4.

Combining these conditions, x must be greater than 1, not equal to 2, and between -4 and 4. So, x ∈ (1, 2) ∪ (2, 4).

d) log(-x)(x²+x-6)

This one's got a negative in the base, which might look scary, but we'll handle it. We have log(-x)(x² + x - 6). Let's go through our conditions:

Base must be positive: -x > 0
- Multiplying by -1 (and flipping the inequality), we get x < 0.
Base cannot be 1: -x ≠ 1
- Multiplying by -1, we get x ≠ -1.
Argument must be positive: x² + x - 6 > 0
- Factoring the quadratic, we get (x + 3)(x - 2) > 0.
- To solve this inequality, we consider the intervals determined by the roots -3 and 2. We find that the inequality holds when x < -3 or x > 2.

Now, let's put it all together. x must be less than 0, not equal to -1, and either less than -3 or greater than 2. Since x has to be less than 0, the only interval that matters is x < -3. So, x ∈ (-∞, -3).

e) log(x-1)

Here we have log(x-1). When no base is written, it's assumed to be base 10. So, we have log₁₀(x-1). Let's apply our rules:

Base is 10, which is positive and not equal to 1, so we're good there.
Argument must be positive: x - 1 > 0
- Adding 1 to both sides, we get x > 1.

So, for this logarithm to be defined, x must be greater than 1. In interval notation, x ∈ (1, ∞).

f) log

This looks incomplete! It seems like there's a missing argument. In this case, we can't determine any values of x because there's nothing to evaluate the logarithm of. We need an argument like log(x) or log(5), etc.

g) log 4

Again, no base is specified, so we assume it's base 10: log₁₀(4). The argument (4) is positive, and the base (10) is positive and not equal to 1. There's no x to solve for here; this logarithm is defined as it is.

h) log(x+4)3

Here we have log(x+4)3. Let's check our conditions:

Base must be positive: x + 4 > 0
- Subtracting 4 from both sides, we get x > -4.
Base cannot be 1: x + 4 ≠ 1
- Subtracting 4 from both sides, we get x ≠ -3.
Argument (3) is already positive, so we're good there.

So, x must be greater than -4, but it can't be -3. In interval notation, x ∈ (-4, -3) ∪ (-3, ∞).

i) log(-x+2)5

Now we have log(-x+2)5. Let's run through our rules:

Base must be positive: -x + 2 > 0
- Adding x to both sides, we get 2 > x, or x < 2.
Base cannot be 1: -x + 2 ≠ 1
- Subtracting 2 from both sides, we get -x ≠ -1.
- Multiplying by -1, we get x ≠ 1.
Argument (5) is already positive, so we're good there.

So, x must be less than 2 and not equal to 1. In interval notation, x ∈ (-∞, 1) ∪ (1, 2).

j) log(2-x)

This one's log(2-x), and we assume the base is 10. Let's see what we need:

Base is 10, which is fine.
Argument must be positive: 2 - x > 0
- Adding x to both sides, we get 2 > x, or x < 2.

So, x must be less than 2. In interval notation, x ∈ (-∞, 2).

k) log(x+3)(x-1)

Here we have log(x+3)(x-1). Let's break it down:

Base must be positive: x + 3 > 0
- Subtracting 3 from both sides, we get x > -3.
Base cannot be 1: x + 3 ≠ 1
- Subtracting 3 from both sides, we get x ≠ -2.
Argument must be positive: x - 1 > 0
- Adding 1 to both sides, we get x > 1.

Combining these, x must be greater than -3, not equal to -2, and greater than 1. The most restrictive condition is x > 1, so x ∈ (1, ∞).

l) log3

Again, there's a missing argument! We can't determine values of x without an argument for the logarithm.

m) log(x-1)(2-5x+4)/(x+1)

Okay, this one's a doozy! We have log(x-1)(2 - 5x + 4) / (x + 1). Let's simplify the argument first:

Argument: (2 - 5x + 4) / (x + 1) = (6 - 5x) / (x + 1)

Now let's apply our rules:

Base must be positive: x - 1 > 0
- Adding 1 to both sides, we get x > 1.
Base cannot be 1: x - 1 ≠ 1
- Adding 1 to both sides, we get x ≠ 2.
Argument must be positive: (6 - 5x) / (x + 1) > 0
- To solve this rational inequality, we find the critical points by setting the numerator and denominator equal to zero:
  - 6 - 5x = 0 → x = 6/5
  - x + 1 = 0 → x = -1
- Now we test intervals around these critical points:
  - For x < -1, both the numerator and denominator are positive, so the fraction is positive.
  - For -1 < x < 6/5, the numerator is positive and the denominator is negative, so the fraction is negative.
  - For x > 6/5, both the numerator and denominator are negative, so the fraction is positive.
- We want the intervals where the fraction is positive, so x < -1 or x > 6/5.

Combining all the conditions:

x > 1
x ≠ 2
x < -1 or x > 6/5

The only interval that satisfies all these conditions is x > 6/5 and x ≠ 2. So, x ∈ (6/5, 2) ∪ (2, ∞).

Final Thoughts

Phew! We tackled a lot of logarithms today, guys! The key takeaway is always to remember those core conditions: the base must be positive and not equal to 1, and the argument must be positive. By carefully applying these rules, you can confidently determine the values of x that make logarithmic expressions defined. Keep practicing, and you'll become a logarithm master in no time!