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SectionProperties of Logarithms
Recall that an exponential function is a function of the form
\begin{equation*}
P=P(t)=a(\alert{b})^t.
\end{equation*}
In this section we will discuss logarithmic functions, which are inverses of exponential functions. In particular, the function
\begin{equation*}
t=\log_{\alert{b}}\left(\frac{P}{a}\right)
\end{equation*}
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:
The Logarithm
Let \(b\neq 1 \) be a positive number, then the function
\begin{equation*}
f(t)=\log_b(t)
\end{equation*}
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
\begin{equation*}
x=\log_b(y)\ \ \ \ \text{and}\ \ \ \ b^x=y
\end{equation*}
are equivalent.
Caution221
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:
\begin{equation*}
\log_{10}(y)=\log(y).
\end{equation*}
In other words, we simply drop the subscript when referring to base \(10 \text{.}\)
SubsectionProperties of Logarithms
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Recall the following laws of exponents:
-
To multiply two powers with the same base, add the exponents and leave the base unchanged.
\begin{equation*} a^m \cdot a^n = a^{m+n} \end{equation*} -
To divide two powers with the same base, subtract the exponents and leave the base unchanged.
\begin{equation*} \frac{a^m}{a^n} = a^{m-n} \end{equation*} -
To raise a power to a power, keep the same base and multiply the exponents.
\begin{equation*} \left(a^m\right)^n = a^{mn} \end{equation*}
Each of these laws corresponds to one of three properties of logarithms.
Properties of Logarithms
If \(x,y,b>0\text{,}\) and \(b\neq 1\text{,}\) then
\(\log_b(xy)=\log_b(x)+\log_b(y),\)
\(\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\)
\(\log_b(x^k)=k\cdot \log_b(x),\)
\(\log_b(b^y)=y,\)
\(b^{\log_b(x)}=x.\)
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 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}\left({\frac{16}{2}}\right)=\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*}
SubsectionUsing the Properties of Logarithms
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 logarithm in terms of simpler logarithms.
Example222
Simplify \(\log_{b}\sqrt{{xy}}\text{.}\)
Solution
First, we write \(\sqrt{xy}\) using a fractional exponent:
\begin{equation*}
\log_{b}{\sqrt{xy}} = \log_{b}{\big((xy)^{1/2}\big)}.
\end{equation*}
Then we apply Property (3) to rewrite the exponent as a coefficient:
\begin{equation*}
\log_{b}{((xy)^{1/2})} = \frac{1}{2}\log_{b}{(xy)}.
\end{equation*}
Finally, by Property (1) we write the log of a product as a sum of logs:
\begin{equation*}
\frac{1}{2}(\log_{b}({xy}) = \frac{1}{2}(\log_{b}({x}) + \log_{b}({y}).
\end{equation*}
Thus, \(\log_{b}\sqrt{xy} = \frac{1}{2}(\log_{b}({x}) + \log_{b}({y})\text{.}\)
Example223
Simplify \(\log_{b}({xy^2})\text{.}\)
Solution
\begin{equation*}
\begin{aligned}
\log_b(xy^2) \amp = \log_b(x)+\log_b(y^2) \amp\amp\text{Property 1} \\
\amp = \log_b(x)+2\log_b(y) \amp\amp\text{Property 3}
\end{aligned}
\end{equation*}
Caution224
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.
Example225
Express \(3(\log_b (x) - \log_b (y))\) as a single logarithm with a coefficient of \(1\text{.}\)
Solution
We begin by applying Property (2) to combine the logs.
\begin{equation*}
3(\log_b (x) - \log_b (y)) = 3 \log_{b}\left(\frac{x}{y}\right)
\end{equation*}
Then, using Property (3), we replace the coefficient \(3\) by an exponent \(3\text{.}\)
\begin{equation*}
3 \log_{b}\left(\frac{x}{y}\right)=\log_{b}\left(\left(\frac{x}{y}\right)^3\right)
\end{equation*}
Example226
Express \(2\log_b (x) + 4\log_{b}{(x + 3)}\) as a single logarithm with a coefficient of \(1\text{.}\)
Answer
\begin{equation*}
\begin{aligned}
2\log_b(x)+4\log_b(x+3) \amp= \log_b(x^2)+\log_b\big((x+3)^4\big) \amp\amp\text{Property 3} \\
\amp= \log_b(x^2(x+3)^4) \amp\amp\text{Property 1}
\end{aligned}
\end{equation*}
SubsectionThe Natural Exponential Function
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
\begin{equation*}
e \approx 2.71828182845.
\end{equation*}
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
The natural logarithm is the logarithm base \(e\text{.}\)
\begin{equation*}
\ln(x) = \log_{e}({x}), ~~~~ x\gt 0
\end{equation*}
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 \(e\text{.}\)
SubsectionSolving Exponential Equations
Suppose we want to solve the equation
\begin{equation*}
5^x = 7.
\end{equation*}
We could rewrite the equation in logarithmic form to obtain the exact solution
\begin{equation*}
x = \log_{5}({7}).
\end{equation*}
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
\begin{equation*}
\log_{10}{(5^x)} = \log_{10}({7}).
\end{equation*}
Then we use Property (3) to rewrite the left side as
\begin{equation*}
x \log_{10}({5}) = \log_{10}({7}).
\end{equation*}
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
\begin{equation*}
x = \frac{\log_{10}({7})}{\log_{10}({5})}.
\end{equation*}
On your calculator, enter the sequence
LOG( \(7\) ) \(\div\) LOG ( \(5\) ) ENTER
to find that \(x \approx 1.2091\text{.}\)
Caution227
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
\begin{equation*}
\dfrac{\log_{10}({7})}{\log_{10}({5})}\ne \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}({1.4})\text{.}
\end{equation*}
Example228
Solve \(1640 = 80 \cdot 6^{0.03x}\text{.}\)
Solution
First we divide both sides by \(80\) to obtain
\begin{equation*}
20.5 = 6^{0.03x}.
\end{equation*}
Next, we take the base \(10\) logarithm of both sides of the equation and use Property (3) of logarithms to get
\begin{equation*}
\log_{10}{20.5} = \log_{10}{(6^{0.03x})}= 0.03x \log_{10}{6}.
\end{equation*}
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:
\begin{equation*}
x = \frac{\log_{10}{20.5}}{0.03 \log_{10}{6}}\approx 56.19.
\end{equation*}
(On your calculator, remember to enclose the denominator, \(0.03 \log_{10}{6}\text{,}\) in parentheses.)
Caution229
In Example228, do not try to simplify
\begin{equation*}
80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \text{ Incorrect!}
\end{equation*}
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by \(80\text{.}\)
Example230
Solve \(5(1.2)^{2.5x} = 77\text{.}\)
Solution
First we divide both sides by \(5\text{.}\)
\begin{equation*}
1.2^{2.5x}=15.4
\end{equation*}
Next, we take the base \(10\) logarithm on both sides and use Property (3):
\begin{equation*}
\begin{aligned}
\log_{10}(1.2^{2.5x}) \amp= \log_{10}(15.4) \\
(2.5x)\log_{10}(1.2) \amp= \log_{10}(15.4)
\end{aligned}
\end{equation*}
Finally, solving for \(x\) we have
\begin{equation*}
\begin{aligned}
(2.5x)\log_{10}(1.2) \amp= \log_{10}(15.4) \\
x \amp= \frac{\log_{10}(15.4)}{2.5\log_{10}(1.2)}\approx 5.999
\end{aligned}
\end{equation*}