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Section 1.5 Inverse Functions and Logarithms

Equation solving and inverse functions.

In part (c) of Example A of the previous section, we knew that the value of the function \(m(t)\) (mass of Strontium-90 remaining) was 5 mg, and wanted to know the corresponding value of its argument \(t\) (the time).

More generally it would be useful to have a formula giving time \(t\) as a function of mass \(m\text{,}\) \(t=g(m)\text{.}\) A function like this that takes values “backwards” compared to the function is the inverse of that function.

For example, with \(y=f(x)=x^{3}\text{,}\) we get back from a \(y\) value to the corresponding \(x\) value by treating \(y\) as known and solving \(y=x^3\) for the unknown \(x\text{:}\) this gives \(x=\sqrt[3]{y}\text{.}\) Thus the cube root function is the inverse of the cube function, and we write \(x=f^{-1}(y)=\sqrt[3]{y}\text{.}\)

It is often convenient to go back to using the name \(x\) for the argument of this new function too, writing \(f^{-1}(x)=\sqrt[3]{x}\text{.}\)

The Horizontal Line Test [HLT] and one-to-one functions.

Graphically, one gets the graph of the inverse function \(x=f^{-1}(y)\) by flipping the graph of \(y=f(x)\) along the diagonal line \(y=x\text{.}\) Since the role of the \(x\) and \(y\) values are swapped, the domain of the inverse is the range of the original function, and vice versa.

However, this flipping does not always give the graph of a function. The graph of any function must pass the vertical line test that no vertical line intersects it more than once, and for the flipped graph to pass, the original graph must have no horizontal line intersects it more than once.

This is the Horizontal Line Test [HLT], and is exactly what is needed for a function to have an inverse. Algebraically, this means that no two different arguments \(x_1\) and \(x_2\) give the same value of the function:

Definition 1.5.1.
Function \(f\) is one-to-one if for any \(x_1 \neq x_2\text{,}\) \(f(x_1) \neq f(x_2)\text{.}\)

Is the function \(f(x) = x^3\) one-to-one? Solution
This passes the HLT, since the function is increasing and so passes through any horizontal line just once and never returns. In fact, any function that is always increasing passes the HLT (same if it is always decreasing)
Is the function \(g(x) = x^2\) one-to-one? Solution
No; this fails the HLT: the horizontal line \(y=1\) intersects for both \(x=1\) and \(x=-1\text{.}\) In fact every line \(y=a\) for positive \(a\) intersects at two \(x\) values, \(\sqrt{a}\) and \(-\sqrt{a}\text{.}\)
Inverse functions.
One-to-one functions are exactly the ones that have inverses:
Definition 1.5.4. (Inverse).
If function \(f\) is one-to-one, if has an inverse, denoted \(f^{-1}\text{,}\) and defined by \(x=f^{-1}(y)\) being given by the solution \(x\) of the equation \(f(x)=y\text{.}\)
If a function is not one-to-one, this equation has several solutions for some values of \(y\text{,}\) so does not determine the value \(x\text{;}\) thus the inverse is not defined.
Domain and range of the inverse.
The domain of \(f^{-1}\) is the range of \(f\) [the “y-values” in the above equation], and the range of \(f^{-1}\) is the domain of \(f\) [the “x-values”].
Caution: changing domain gives a different function.

How do we reconcile \(f(x)=x^2\) not being one-to-one with it having an inverse, the square root function?

Be careful: we are talking about two different functions here, even though they are described using the same formula \(y=x^2\text{!}\) When we use the formula with domain all the real numbers, it is not one-to-one, and has no inverse, but when we change the domain to non-negative real numbers, that is a different function, with inverse \(\sqrt{x}\text{.}\) The difference with this smaller domain is that the graph is only the right half of a parabola, which is increasing and so satisfies the HLT.

Algebraically, for any given value \(y\text{,}\) the equation \(x^2=y\) has only one non-negative solution \(x\text{.}\)

If \(f(1) = 5\text{,}\) \(f(3) = 7\) and \(f(8) = -10\text{,}\) find \(f^{-1}(-10)\text{,}\) \(f^{-1}(5)\) and \(f^{-1}(7)\text{.}\)

Notation warning: inverses are not reciprocals!
Beware of a possible confusion:
\(y=f^{-1}(x)\) [the inverse of function \(f\) applied to \(x\)]
is not the same as
\(y=[f(x)]^{-1} = \frac{1}{f(x)}\) [the reciprocal of \(f(x)\)].
Find the inverse function of \(f(x) = x^3 + 2\text{.}\)
Logarithmic Functions.

Does an exponential function \(y=f(x)=a^{x}\) have an inverse?

For \(a>1\text{,}\) the value of \(a^x\) increases as \(x\) increases: the graph is increasing, which is enough to pass the HLT and ensure existence of an inverse.

For \(0 \lt a \lt 1\text{,}\) the graph is decreasing and so again passes the HLT, giving an inverse. (For \(a=1\text{,}\) there is no inverse, and the function is boring: \(f(x) = 1\text{.}\))

This inverse should be familiar: the number \(x\) for which \(a^x=y\) is called the logarithm of \(y\) base \(a\), written \(\log_{a}y\text{,}\) so the inverse of the exponential function \(f(x)=a^{x}\) is the logarithmic function base \(a\), \(f^{-1}(x)=\log_{a}x\text{.}\)

An exponential function for \(a \neq 1\) is defined for all real numbers (its domain) and its values (range) are all positive numbers. Thus the logarithmic functions \(\log_{a}\) have domain all the positive numbers, range all the reals: only positive numbers have logarithms.

Note that this simple domain and range for logarithms depends on exponential functions being defined for all real arguments, not just all rational arguments.

Rules for logarithms.
Logarithms satisfy the following rules, all following from the rules for exponentials in Section 1.5; For any positive number \(a\) except 1, and any positive numbers \(x\) and \(y\text{,}\)
  1. \(\displaystyle \log_a(x \cdot y) = \log_a x + \log_a y\)
  2. \(\displaystyle \log_a(x/y) = \log_a x - \log_a y\)
  3. \(\log_a(x^p)=p \log_a x\) for any real power \(p\text{.}\)
Natural logarithms.

Since \(e^{x}\) is the most commonly used exponential function, its inverse \(\log_{e}\) is the most important logarithmic function: it is called the natural logarithm, and has the special name \(\ln\) (from the initials of “logarithm” and “natural”):

\begin{equation*} \ln x = \log_e x. \end{equation*}

Our first use of the natural logarithm is to put any exponential function \(a^{x}\) in terms of \(e^{x}\text{.}\) Using the properties of exponentials and the fact that \(e^{\ln a}=a\text{,}\)

\begin{equation*} a^x = (e^{\ln a})^x = e^{(\ln a)x}. \end{equation*}

It can also be shown that

\begin{equation*} \log_{a} x = \frac{\ln x}{\ln a}, \end{equation*}

so we can also put all logarithmic functions in terms of natural logarithms.

Thus we mostly need just one exponential function, \(e^{x}\text{,}\) and just one logarithmic one: its inverse, the natural logarithm.

Recommended Exercises Recommended Exercises

Study Exercises 3, 4, 6, 10, 13, 17, 18, 19, 22, 25 from the text.

We omit the final topic, Inverse trigonometric functions, for now; instead we will review it when we encounter these functions in Section 3.5.