\(\log _A(B)\) is pronounced “the logarithm base \(A\) of \(B\)” or “the log base \(A\) of \(B\)” or “the logarithm of \(B\) base \(A\)”.

Logarithms are exponents. They are the things we raise bases to, to get targeted outputs. Therefore, to understand logarithms, we should investigate the exponents in exponential functions.

An example of a basic exponential function is \(E(t) = 2^t\).

Its graph looks like

\(E(t)\) is an increasing function, which means when we select a greater exponent in the domain, then we get a greater function value.

We can see this by evaluating our exponential function with increasing domain numbers:

• \(E(-1) = 2^{-1} = \frac {1}{2}\)
• \(E(0) = 2^0 = 1\)
• \(E\left (\frac {1}{2}\right ) = 2^{\tfrac {1}{2}} = \sqrt {2}\)

When we evaluate this exponential function, we know the domain number and we seek its range partner. In this case, we select \(d\) from the domain and we assemble the pair \((d, 2^d)\), where \(2^d\) is the value of the function, \(E(t)\), at \(d\).

Here we are supplying an exponent from the domain and then obtaining a function value for the exponential function. We can look at this process in reverse: we select a function value and we are in search of an exponent that will produce the desired output.

Reverse

How do we think about \(E(t) = 2^t\) in reverse?

• \(E(t) = {\tfrac {1}{4}} \)
• \(E(t) = 4\)
• \(E(t) = 16 \)

Here, we know the desired value of the function. We seek the domain numbers paired with it. We seek the exponent. We know \(2^d\), we seek the pair \((d, 2^d)\), where \(d\) is the solution to the equation \(E(t) = 2^d\).

\(\bigstar \) Since \(E(t) = 2^t > 0\) for any \(t\), thinking backwards only works if we provide positive function values. \(2^t\) cannot equal a negative value. \(2^t\) cannot equal \(0\).

\(\bigstar \) For \(E(t) = 2^t\), every positive function value has exactly one associated domain number, one exponent, one logarithm. (That is eeriely similar to the function rule for pairs in the exponential function.)

Thinking of an exponential function in reverse sounds like a new function. We call it a logarithmic function.

• In reverse, the range of an exponential function becomes the domain of a logarithmic function.
• In reverse, the domain of an exponential function becomes the range of a logarithmic function.

The basic logarithmic function just reverses the pairs in the basic exponential function. If \((a, 2^a)\) is a pair in the exponential function, then \((2^a, a)\) is a pair in the logarithmic function.

Therefore, the domains and ranges switch.

Logarithm

It seems weird to write \((2^a, a)\), even though it is perfectly correct. We are supplying values of an exponential function, which look like \(2^a\) and then the value of the logarithmic function is the needed exponent.

Instead, we are used to writing the domain number by a letter, like \(b\), and then the function value as a formula involving \(b\). That’s what we are used to.

We would prefer our pairs in the logarithm function to look like

Logarithmic functions come from the study of logarithms, which is the study of the exponents, just like we are talking about. This gets shortened to log for notational purposes.

In our example here, we are working with the base \(2\) exponential function, \(E(t) = 2^t\). So, the reverse is called the logarithm base \(2\). We tack on a subscript \(2\) to the log to remind us of the base.

With this new symbol, we can describe pairs in the logarithmic function and points on its graph like

\(\log _2(b)\) is just the exponent that \(2\) needs to equal \(b\).

\(\blacktriangleright \) Remember: These pairs, \((b, \log _2(b))\), are the reverse of the exponential pairs : \((a, 2^a)\). Therefore, \((b, \log _2(b))\) is also \((2^c, c)\) for some \(c\).

\(\blacktriangleright \) \(\log _2(b)\) is the exponent, so that \(2^? = b\)

\(\blacktriangleright \) \(\log _2(b)\) is the number that you raise \(2\) to, to get \(b\).

We have a logarithmic function for every exponential function. They are designated by their bases.

Graphs

Since all of the pairs of exponential and logarithmic functions are reversed, the graphs switch axes.

The horizontal axis is an asymptote for the graph of a basic exponential function. Switching axes makes the vertical axis an asymptote for the graph of a basic logarithmic function.

The graph of a basic exponential function has \((0,1)\) as an intercept. The graph of a basic logarithmic function has \((1,0)\) as an intercept.

Their graphs are mirror images of each other across the diagonal through quandrants I and III.

From this basic logarithmic function and its graph we can transform and analyze more general logarithmic functions.

Note: Many times people just speak about exponential functions, and they mean both exponential and shifted exponential.

In this example, \(M(t)\) is a logarithmic function, so it must have an inverse partner exponential function. The pairs for \(M\) look like \((t, M(t))\) or just \((t,M)\). The pairs for the exponential function would look like \((M, t)\). The roles of \(M\) and \(t\) would be switched. \(M\) would be the variable in the formula. \(t\) would be the function value.

\(M = \log _2(t+3) - 4\) is an equation describing both the logarithmic function, \(M(t)\), and the exponential function \(t(M)\). We just are used to having the function name by itself on one side of the equal sign.

We can obtain such a formula for this partner exponential function by solving the logarithmic equation for \(t\).

\begin{align*} \log _2(t+3) - 4 & = M \\ \log _2(t+3) & = M + 4 \\ 2^{\log _2(t+3)} & = 2^{M+4} \\ t+3 & = 2^{M+4} \\ t & = 2^{M+4} - 3 \end{align*}

\(\blacktriangleright \) Remember: \(\log _2(t+3)\) is the thing that you raise \(2\) to, to get \(t+3\) and \(\log _2(t+3) = M+4\). Therefore, \(M+4\) is the thing that you raise \(2\) to, to get \(t+3\)

Here is the graph of both the logarithmic and the associated exponential functions. They are inverses of each other.

Reversing all of the pairs will give the inverse exponential function.

\(K(x)\) is a logarithmic function, so it must have a partner exponential function. The pairs for \(K\) look like \((x, K)\). The pairs for the exponential function would look like \((K, x)\). The roles of \(K\) and \(x\) would be switched. \(K\) would be the variable in the exponential formula. \(x\) would be the exponential function value. \(K\) would be the independent variable and \(x\) would be the dependent variable.

We can obtain a nice formula for this partner exponential function by solving the logarithmic equation for \(x\).

\begin{align*} -2 \, \log _3(4-x) & = M \\ \log _3(4-x) & = -\frac {M}{2} \\ 3^{\log _3(4-x)} & = 3^{-\frac {M}{2}} \\ 4 - x & = 3^{-\frac {M}{2}} \\ 4 - 3^{-\frac {M}{2}} & = x \\ \end{align*}

\(\blacktriangleright \) Remember: \(\log _3(4-x)\) is the thing that you raise \(3\) to, to get \(4-x\) and \(\log _3(4-x) = - \frac {M}{2}\). Therefore, \(- \frac {M}{2}\) is the thing that you raise \(3\) to, to get \(4-x\).

Here is the graph of both the logarithmic and exponential functions. They are reflected about the diagonal, because the functions are inverses of each other.

2025-08-05 23:50:00