## 17Calculus Precalculus - Domain and Range of Logarithms

##### 17Calculus

Domain and Range of Logarithms

For this discussion, we assume that the base of the logarithm is positive. This is a common assumption but it is not always stated in textbooks and by instructors. We will use the natural logarithm here since you will use the natural log almost exclusively in calculus.

If you think about the relationship between exponents and logarithms, it looks like this.

 $$x=e^y \to \ln(x) = y$$

Since $$e \approx 2.71828$$, $$e$$ is positive. So taking $$e$$ to any power $$y$$ always results in a positive number $$x$$. Therefore going in the other direction, $$x$$ must always be positive. Do you see that?
Now, what about zero? Can we take the logarithm of zero? Well, to answer that, let's look back at $$x=e^y$$. If we want $$x$$ to be zero, is there any number that we can choose for the exponent $$y$$ to get $$e^y = 0$$?
The answer is no. Even if we try very, very large negative numbers for $$y$$, $$x$$ will always be greater than zero. So in the logarithm equation, $$x$$ can never equal zero.

So, for our domain of the logarithm, we have

Domain of $$y=\ln(x)$$

$$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 0 \}$$

Just a note on notation here. This set notation for the domain says, $$x$$ must be a real number, $$x ~ \in \mathbb{R}$$ and $$x$$ is greater than zero, $$x \gt 0$$. This notation is pretty formal. So you might see something like this $$D: x ~\in (0,\infty)$$ which implies that $$x$$ must be a real number without actually stating it. The capital 'D' stands for domain.

For the range, we have no restrictions on what we can use as the exponent $$y$$ in $$x=e^y$$, giving us $$\{ y ~|~ y ~ \in \mathbb{R} \}$$ as the range in $$y=\ln(x)$$. This can be written as $$R: y ~\in (-\infty, +\infty)$$ where the capital 'R' stands for range.

Okay, let's try some practice problems.

Practice

Unless otherwise instructed, determine the domain of these logarithmic functions.

$$f(x) = \ln(x-3)$$

Problem Statement

Find the domain of the logarithm function $$f(x) = \ln(x-3)$$

The domain of $$f(x) = \ln(x-3)$$ is $$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 3 \}$$

Problem Statement

Find the domain of the logarithm function $$f(x) = \ln(x-3)$$

Solution

Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is
$$x-3 \gt 0$$. Solving for $$x$$, we get $$x \gt 3$$.

The domain of $$f(x) = \ln(x-3)$$ is $$\{ x ~|~ x ~ \in \mathbb{R} ~ and ~ x \gt 3 \}$$

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$$F(x) = \log_2(x^2)$$

Problem Statement

Find the domain of the logarithm function $$F(x) = \log_2(x^2)$$

The domain of $$F(x) = \log_2(x^2)$$ is $$x \neq 0$$, i.e. all real numbers except for $$x=0$$.

Problem Statement

Find the domain of the logarithm function $$F(x) = \log_2(x^2)$$

Solution

Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$x^2 \gt 0$$. Since $$x$$ is squared, it is always positive except when $$x=0$$. So $$x$$ can take on any value as long as it is not zero.

The domain of $$F(x) = \log_2(x^2)$$ is $$x \neq 0$$, i.e. all real numbers except for $$x=0$$.

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$$y = \log_2(5-x) + 3$$

Problem Statement

Find the domain of the logarithm function $$y = \log_2(5-x) + 3$$

Solution

### 3031 video solution

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$$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$

The domain of $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$ is all real numbers that satisfy $$x \gt 10$$.

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$

Solution

Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$x/2 - 5 \gt 0$$. No other term in the function affects the domain. Solving for $$x$$, we have $$x/2 \gt 5 \to x \gt 10$$.

The domain of $$\displaystyle{ f(x) = 3 - 2\log_4 \left( \frac{x}{2} - 5 \right) }$$ is all real numbers that satisfy $$x \gt 10$$.

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$$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$

The domain of $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$ is $$x \gt -1$$.

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$

Solution

Since we cannot take the logarithm of non-positive (zero and negative) numbers, we need the expression inside the natural logarithm to be greater than zero. In equation form, this is $$\displaystyle{ \frac{1}{x+1} \gt 0 }$$. Since the numerator is always positive, we also need the denominator to be positive, i.e. $$x+1 \gt 0 \to x \gt -1$$. Looking back at the original function, we also have to make sure that the denominator of the fraction inside the logarithm is never zero. This is taken care of as long as $$x \neq -1$$. We are already including that in the inequality $$x \gt -1$$, so no more work needs to be done.

The domain of $$\displaystyle{ f(x) = \ln \left( \frac{1}{x+1} \right) }$$ is $$x \gt -1$$.

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$$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$

The domain of $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$ is $$x \gt 0$$ OR $$x \lt -1$$. This can also be stated as all real numbers except for $$-1 \leq x \leq 0$$.

Problem Statement

Find the domain of the logarithm function $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$

Solution

For this one, let's start with the rational expression inside the logarithm. We need to make sure we never have a zero in the denominator. So $$x \neq 0$$. Now need to make sure that the expression inside the logarithm is always positive. For rational expressions, the numerator and denominator need to be of the same sign, i.e. they are either both positive or both negative.
So for both to positive, we need $$x+1 \gt 0 \to x \gt -1$$ AND $$x \gt 0$$. The inequality $$x \gt 0$$ covers both cases.
Now for both to be negative, we need $$x+1 \lt 0 \to x \lt -1$$ AND $$x \lt 0$$. The first inequality, $$x \lt -1$$ covers both cases.
Now let's put all three cases together.

$$x \neq 0$$

$$x \gt 0$$

$$x \lt -1$$

Okay, so the first statement is covered in both of the other two, so we can ignore it. Combining the last two inequalities gives us the domain.

The domain of $$\displaystyle{ g(x) = \log_5 \left( \frac{x+1}{x} \right) }$$ is $$x \gt 0$$ OR $$x \lt -1$$. This can also be stated as all real numbers except for $$-1 \leq x \leq 0$$.

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$$f(x) = \sqrt{\ln x}$$

Problem Statement

Find the domain of the logarithm function $$f(x) = \sqrt{\ln x}$$

The domain of $$f(x) = \sqrt{\ln x}$$ is $$x \geq 1$$.

Problem Statement

Find the domain of the logarithm function $$f(x) = \sqrt{\ln x}$$

Solution

Let's start with the logarithm. Since we can't take logarithms of negative numbers, we need $$x \gt 0$$. So that was pretty easy. Now we want to make sure that we don't take the square root of negative numbers. So we need $$\ln(x) > 0$$. Can this ever happen? Yes, it can since the range of a logarithm is all real numbers. So when will the logarithm be negative? By looking at a graph, we know that $$\ln 1 = 0$$ and values of $$x$$ less than $$1$$ produce negative values for the logarithm. So we need $$x \geq 1$$. Notice here that it is okay to take the square root of zero, so we include $$x = 1$$.
Okay, so putting the two inequalities together, we have $$x \gt 0$$ and $$x \geq 1$$. The first one is covered by the second one, so we just need the second one as our final answer.

The domain of $$f(x) = \sqrt{\ln x}$$ is $$x \geq 1$$.

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