From Exponential Functions to Logarithms
Exponential functions answer questions of the form: if a quantity starts at a given value and is multiplied by the same factor repeatedly, what value results after a specified number of steps? A logarithm answers the reverse question: which exponent produces a specified positive value?
For example, the exponential function \(f(x)=2^x\) takes an exponent as its input and produces a positive output. Its values include \(2^3=8\). The inverse function reverses this input-output relationship: it takes \(8\) as its input and returns the exponent \(3\). This inverse function is the logarithm with base \(2\).
To make that reversal possible for every positive input, the exponential function must have an inverse whose domain is all positive real numbers. The choice of base is therefore important.
Exponential Functions and Their Domains and Ranges
For a fixed base \(b>0\), the exponential function is \(f_b:\mathbb{R}\to(0,\infty)\), defined by \(f_b(t)=b^t\). Its domain is all real numbers: any real number may be used as an exponent. Its range is the positive real numbers, since \(b^t>0\) for every real \(t\).
When \(b>1\), the exponential function is strictly increasing. When \(0<b<1\), it is strictly decreasing. In either case it is one-to-one, so distinct inputs give distinct outputs. Its range includes every positive real number. These properties mean that \(f_b\) is a bijection from \(\mathbb{R}\) to \((0,\infty)\), and therefore has an inverse function.
The input to \(\log_b\) must be positive because the range of \(f_b\) consists exactly of positive numbers. The output of \(\log_b\) can be any real number because every real number is an allowed input to \(f_b\). The notation \(\log_b(x)\) names a function value; \(b\) is its base and \(x\) is its input.
Why the Base Must Be Positive and Not One
The requirement \(b>0\) ensures that \(b^t\) is a positive real number for every real exponent \(t\). A negative base does not give real values for all real exponents. For instance, \((-2)^{1/2}\) is not a real number, so \(t\mapsto(-2)^t\) cannot serve as a real-valued function on all of \(\mathbb{R}\).
The requirement \(b\ne1\) ensures that the exponential function is one-to-one and has every positive real number in its range. If \(b=1\), then \(1^t=1\) for every real \(t\). The function is constant: it never produces, for example, \(2\), and many different inputs produce the same output. It therefore has no inverse function from \((0,\infty)\) to \(\mathbb{R}\).
Thus, the allowed bases are exactly those satisfying \(b>0\) and \(b\ne1\). They fall into two cases: \(b>1\), which gives exponential growth, and \(0<b<1\), which gives exponential decay. Both cases produce valid logarithmic inverse functions.
The Inverse Relationship
An inverse function reverses the action of the original function. If a function \(f:A\to B\) is a bijection, its inverse \(f^{-1}:B\to A\) undoes \(f\): applying \(f\) and then \(f^{-1}\) returns the original input, and applying \(f^{-1}\) and then \(f\) does the same.
These identities follow directly from the definition of an inverse. The first starts with an exponent \(t\), applies the exponential function, and then applies its inverse. The second starts with a positive value \(x\), applies the inverse, and then applies the exponential function. In each case, the two functions undo one another.
Why the Domains and Ranges Switch
The domain of a function is the set of inputs it accepts, and its range is the set of outputs it actually produces. An inverse reverses every input-output pair. Consequently, the original range becomes the inverse domain, and the original domain becomes the inverse range.
\(f_b\) accepts every real input \(t\) and produces a positive output \(b^t\). Its domain is \(\mathbb{R}\), and its range is \((0,\infty)\).
The inverse accepts an output of \(f_b\) as its input. Therefore, \(\log_b\) accepts positive inputs, the range of \(f_b\).
The output of \(\log_b\) is the exponent used by \(f_b\). Every real number is such an exponent, so the range of \(\log_b\) is \(\mathbb{R}\).
This reasoning establishes the domain and range without needing to calculate particular logarithm values. The base determines which exponential function is being inverted; the inverse relationship determines how its domain and range are exchanged.
| Function | Domain | Range |
|---|---|---|
| \(f_b(t)=b^t\), for \(b>0\), \(b\ne1\) | \(\mathbb{R}\) | \((0,\infty)\) |
| \(\log_b(x)\), for \(b>0\), \(b\ne1\) | \((0,\infty)\) | \(\mathbb{R}\) |
Worked Example: The Base-Two Logarithm
Worked Example: The Base-Two Logarithm
Let \(f(t)=2^t\), and let \(g\) be its inverse. Identify the domain and range of each function, and find \(g(8)\).
Step 1: Check the base. The base is \(2\), so \(2>0\) and \(2\ne1\). The inverse function is therefore defined.
Step 2: State the exponential function's domain and range. The input \(t\) may be any real number, so the domain of \(f\) is \(\mathbb{R}\). Every value \(2^t\) is positive, and the exponential function takes every positive value. Thus, the range is \((0,\infty)\).
Step 3: Exchange domain and range for the inverse. The inverse is \(g(x)=\log_2(x)\). Its domain is \((0,\infty)\), and its range is \(\mathbb{R}\).
Step 4: Find \(g(8)\) and check. Since \(2^3=8\), the inverse returns the exponent \(3\), so \(g(8)=\log_2(8)=3\). Check by applying \(f\): \(f(3)=2^3=8\), the original input to \(g\). The result is verified.
Worked Example: A Base Between Zero and One
Worked Example: A Base Between Zero and One
Let \(f(t)=(1/3)^t\), and let \(g\) be its inverse. State the domain and range of each function, then find \(g(1/9)\) and \(g(3)\).
Step 1: Check the base. Here \(b=1/3\), so \(b>0\) and \(b\ne1\). Although \(b<1\), this is a valid base. The exponential function is decreasing, but it is still one-to-one and has range \((0,\infty)\).
Step 2: Identify the domains and ranges. The domain of \(f\) is \(\mathbb{R}\), and its range is \((0,\infty)\). The inverse is \(g(x)=\log_{1/3}(x)\), so its domain is \((0,\infty)\), and its range is \(\mathbb{R}\).
Step 3: Find \(g(1/9)\). We have \((1/3)^2=1/9\), so \(g(1/9)=2\). Check: \(f(2)=(1/3)^2=1/9\), which returns the input.
Step 4: Find \(g(3)\). A negative exponent gives \((1/3)^{-1}=3\). Therefore \(g(3)=-1\). Check: \(f(-1)=(1/3)^{-1}=3\), again returning the input.
The output \(-1\) is permitted because the range of the logarithmic function is all real numbers. A positive input to the logarithm does not require a positive output.
Worked Example: A General Base and Input
Worked Example: A General Base and Input
Let \(h(x)=\log_5(x)\). Determine its domain and range, and find \(h(1/25)\).
Step 1: Check the base. The base \(5\) satisfies \(5>0\) and \(5\ne1\), so \(h\) is defined as the inverse of \(t\mapsto5^t\).
Step 2: Identify the domain and range. The exponential function \(5^t\) has domain \(\mathbb{R}\) and range \((0,\infty)\). Reversing its input-output pairs shows that \(h\) has domain \((0,\infty)\) and range \(\mathbb{R}\).
Step 3: Check whether the proposed input is allowed. Since \(1/25>0\), the input belongs to the domain of \(h\).
Step 4: Find and verify the value. We have \(5^{-2}=1/25\), so \(h(1/25)=-2\). Check using the inverse relationship: \(5^{h(1/25)}=5^{-2}=1/25\), which matches the original input. Thus the value is verified.
Common Mistake: Reversing the Domain and Range Incorrectly
A frequent error is to say that a logarithm accepts every real input because its output can be any real number. This confuses the domain with the range. For \(\log_2(x)\), the input must be in the range of \(2^t\), which is \((0,\infty)\). The output is the exponent, which may be any real number.
For example, \(\log_2(-4)\) is not defined as a real number: \(-4\) is not a possible output of \(2^t\), since \(2^t>0\) for every real \(t\). By contrast, \(\log_2(1/4)\) is defined, even though its value is negative: \(2^{-2}=1/4\), so \(\log_2(1/4)=-2\).
Another mistake is to allow base \(1\). The expression \(1^t=1\) for every \(t\) cannot be reversed to recover a unique exponent. The correction is to check the base before using a logarithm: it must be positive and different from one.
Check Your Understanding
Use the inverse-function definition and the domain and range relationship to answer these questions.
- What are the domain and range of \(x\mapsto 4^x\), and what are the domain and range of its inverse?
- Why does a logarithmic function with base \(1\) not have an inverse on \((0,\infty)\)?
- Is \(\log_{1/2}(8)\) defined as a real number? State the domain condition that decides.
- Given \((1/3)^{-2}=9\), what value must \(\log_{1/3}(9)\) have? How can you check it?
- Explain why a logarithm can have a negative output even though its input must be positive.
