The Logarithm
Let \(b\neq 1 \) be a positive number, then the function
is called a logarithm with base \(b \text{.}\)
Upon inputting a value \(t\) the function \(\log_b(t)\) will tell you the power of \(b \) which will yield \(t \text{.}\)
Recall that in chapter 7 we defined an exponential function to be a function of the form
In this section we will discuss logarithmic functions which are inverses of exponential functions. In particular, the function
is the inverse of the exponential function above. For \(a=1\text{,}\) the following applet illustrates this inverse relationship:
While this inverse relationship is how we will think of logarithms in practice, let's give a formal definition:
Let \(b\neq 1 \) be a positive number, then the function
is called a logarithm with base \(b \text{.}\)
Upon inputting a value \(t\) the function \(\log_b(t)\) will tell you the power of \(b \) which will yield \(t \text{.}\)
Due to the relationship between logarithms and exponentials, we often say that the equations
are equivalent.
One of the more common bases of a logarithm is base \(10 \text{.}\) Since it is used so often, we have developed a short hand notation for a logarithm of base \(10\text{.}\) This short hand is shown below:
In other words, we simply drop the subscript when referring to base \(10 \text{.}\)
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.
To multiply two powers with the same base, add the exponents and leave the base unchanged.
To divide two powers with the same base, subtract the exponents and leave the base unchanged.
To raise a power to a power, keep the same base and multiply the exponents.
Each of these laws corresponds to one of three properties of logarithms.
If \(x,y,b>0\text{,}\) and \(b\neq 1\text{,}\) then
\(\log_b(xy)=\log_b(x)+\log_b(y),\)
\(\log_b\frac{x}{y}=\log_b(x)-\log_b(y),\)
\(\log_b(x^k)=k\cdot \log_b(x),\)
\(\log_b(b^y)=y,\)
\(b^{\log_b(x)}=x.\)
We will examine the properties of logarithms closer in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is is the inverse function of an exponential function.
Property (1): \begin{align*} \log_{2}{32}=\log_{2}{(4 \cdot 8)}=\log_2 4 +\log_2 8= \alert{2} + \alert{3}=\alert{5} \\ \end{align*}
Property (2): \begin{align*} \log_{2}{\frac{16}{2}}=\log_2{16}-\log_2 2=\alert{4}-\alert{1}=\alert{3} \\ \end{align*}
Property (3): \begin{align*} \log_{2}{64}=\log_{2}{(4^3)}=3\log_2 4= \alert{3} \cdot \alert{2}= \alert{6} \\ \end{align*}
Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In the next Example, we rewrite one log in terms of simpler logs.
Simplify \(\log_{b}\sqrt{{xy}}\text{.}\)
First, we write \(\sqrt{xy}\) using a fractional exponent:
Then we apply Property (3) to rewrite the exponent as a coefficient:
Finally, by Property (1) we write the log of a product as a sum of logs:
Thus, \(\log_{b}\sqrt{xy} = \frac{1}{2}(\log_{b}{x} + \log_{b}{y})\text{.}\)
Simplify \(\log_{b}{xy^2}\text{.}\)
Be careful when using the properties of logarithms! Compare the statements below:
\(\log_{b}{(2x)} = \log_{b}{2} + \log_{b}{x} \ \ \ \text{ by Property 1,}\)
but
\(\log_{b}{(2 + x)} \ne \log_{b}{2} + \log_{b}{x}.\)
\(\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b x - \log_b 5 \ \ \ \text{ by Property 2,}\)
but
\(\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b x}{\log_b 5}.\)
We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.
Express \(3(\log_b x - \log_b y)\) as a single logarithm with a coefficient of \(1\text{.}\)
We begin by applying Property (2) to combine the logs.
Then, using Property (3), we replace the coefficient \(3\) by an exponent \(3\text{.}\)
Express \(2\log_b x + 4\log_{b}{(x + 3)}\) as a single logarithm with a coefficient of \(1\text{.}\)
There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called \(e\text{,}\) where
The number \(e\) is essential for many advanced topics, and it is often called the natural base.
The base \(e\) logarithm of a number \(x\text{,}\) or \(\log_ e x\text{,}\) is called the natural logarithm of \(x\) and is denoted by \(\ln x\text{.}\)
The natural logarithm is the logarithm base \(e\text{.}\)
We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied above also apply to logarithms base \(e\text{.}\)
Suppose we want to solve the equation
We could rewrite the equation in logarithmic form to obtain the exact solution
However, sometimes we are stuck in a situation where we cannot evaluate \(\log_{5}{7}\text{.}\) For example, some calculators do not have a log base \(5\) button. So, if we want a decimal approximation for the solution, we begin by taking the base \(10\) logarithm of both sides of the original equation, even though the base of the power is not \(10\text{.}\) This gives us
Then we use Property (3) to rewrite the left side as
Note how using Property (3) allows us to solve the equation: The variable, \(x\text{,}\) is no longer in the exponent, and it is multiplied by a constant, \(\log_{10}{5}\text{.}\) To finish the solution, we divide both sides by \(\log_{10}{5}\) to get
On your calculator, enter the sequence
LOG
(
\(7\) )
\(\div\) LOG
(
\(5\) )
ENTER
to find that \(x \approx 1.2091\text{.}\)
Do not confuse the expression \(\dfrac{\log_{10}{7}}{\log_{10}{5}}\) with \(\log_{10}{\left(\dfrac{7}{5}\right)}\text{;}\) they are not the same! Property (2) allows us to simplify \(\log{\left(\dfrac{x}{y}\right)}\text{,}\) but not \(\dfrac{\log x}{\log y}\text{.}\) We cannot rewrite \(\dfrac{\log_{10}{7}}{\log_{10}{5}}\text{,}\) so we must evaluate it as \((\log 7)/(\log 5)\text{.}\) You can check on your calculator that
Solve \(1640 = 80 \cdot 6^{0.03x}\text{.}\)
First we divide both sides by \(80\) to obtain
Next, we take the base \(10\) logarithm of both sides of the equation and use Property (3) of logarithms to get
On the right side of the equation, \(x\) is multiplied by two constants, \(0.03\) and \(\log_{10}{6}\text{.}\) So, to solve for \(x\) we must divide both sides of the equation by \(0.03 \log_{10}{6}\text{.}\) We use a calculator to evaluate the answer:
(On your calculator, remember to enclose the denominator, \(0.03 \log_{10}{6}\text{,}\) in parentheses.)
In Example216, do not try to simplify
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by \(80\text{.}\)
Solve \(5(1.2)^{2.5x} = 77\text{.}\)
Divide both sides by 5.
Take the log of both sides.
Apply Property (3) to simplify the left side.
Solve for \(x\text{.}\)