2. Logarithms and Exponential Functions and Equations
Equations using logarithms and exponentials allow us to model growth and decay. Rates of growth and decay are relevant to finance, epidemiology, and many areas of science. The laws for working with logarithms enable us to solve equations that cannot be solved with other algebraic techniques.
This chapter:
- Defines logarithms
- Explains the log rules
- Introduces the Log Function
- Introduces the Exponential Function
- Shows how to solve equations involving logs and exponentials.
Do you need this chapter?
Quiz: Logarithms
1. Without using a calculator, determine the numerical values of the following:
[latex]\begin{align*} & \begin{array}{ccc} \textnormal{a) }\log_{2}16 & \textnormal{b) }\log_{10}\sqrt{10} & \textnormal{c) }\log_{10}1000 & \textnormal{d) }\log_{10}\dfrac{1}{100} \end{array} \qquad \qquad \end{align*}[/latex]
Answers
[latex]\begin{array}{lll} \quad \textnormal{a) }4 & \textnormal{b) }\dfrac{1}{2} & \textnormal{c) } 3 & \textnormal{d) } -2 \end{array}[/latex]
2. Simplify the following:
[latex]\begin{align*} \qquad&\text{a) }\log_{4}16+\log_{4}3-\log_{4}6 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad&\text{b) }\frac{1}{2}\log_{10}15-\log_{10}4+2\log_{10}5 \\ \qquad&\text{c) }3\log_{10}2+3\log_{10}5 \\ \qquad&\text{d) }\log_{a}4+2\log_{a}5-2\log_{a}6 \\ \qquad&\text{e) }\frac{1}{2}\log_{10}a^{2}+4\log_{10}b-\log_{10}5ab^{2}\\ \end{align*}[/latex]
Answers
[latex]\begin{array}{lll} &\text{a) }\log_{4}8\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\\ &\text{b) }\log_{10}\left( \frac{25\sqrt{15}}{4}\right) \\ &\text{c) }3\\ &\text{d) }\log_{a}\left( \frac{25}{9} \right)\\ &\text{e) }\log_{10}\left( \frac{b^2}{5} \right)\\ \end{array}[/latex]
3. Simplify the following:
[latex]\begin{align*} &\text{a) }\log_{e}e \\ &\text{b) }\log_{e}e^2 \\ &\text{c) }\log_{e}e^{-2} \\ &\text{d) }\ \log_{e}e^{2}+3\log_{e}2-\log_{e}18 \\ &\text{e) } e^{-\log_{e}\left(2 \right)}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \\ \end{align*}[/latex]
Answers
[latex]\begin{array}{lll} &\text{a) }1\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ &\text{b) }2 \\ &\text{c) }-2\\ &\text{d) }2+\log_{e}\left( \frac{4}{9} \right)\\ &\text{e) } \frac{1}{2}\\ \end{array}[/latex]
4. Solve for [latex]x[/latex] if [latex]\log_{10}(2x-1)=\log_{10}3[/latex]
Answer
[latex]\quad x=2[/latex]
5. Solve for [latex]x[/latex] if [latex]2^{x-4}=12[/latex]. You will need a calculator.
Answer
[latex]\quad x\approx 7.5850[/latex]
The following video provides a quick introduction to logarithms.
Logarithms (6:34 min)
Logarithms (6:34 min) by RMIT University Library Videos (YouTube)
Definition of the logarithm
The logarithm is the inverse of the exponentiation function. Exponentiation is the operation of raising a number called the base to an index (or power) to give a number. The logarithm, in a particular base, of a number is the index (or power) to which the base must be raised to produce the number. For example, the equation
[latex]\begin{align*} 8 & =2^3 \end{align*}[/latex]
means "eight is equal to two to the power of three". Here, 8 is the number, 2 is the base and 3 is the index. This equation has an equivalent logarithmic form and may be written as:
[latex]\begin{align*} \log _2 \left( 8 \right) & =\log_2 \left(2^3 \right)\\ \end{align*}[/latex]
This says "the logarithm to base 2 of 8 is equal to 3". Sometimes, the word logarithm is abbreviated to log and so we would say "the log to base 2 of 8 is equal to 3".
Generally, providing [latex]a>0,\ n>0[/latex] and [latex]a\neq1[/latex],
\[
a^{x}=n \Leftrightarrow \log_{a}n=x.
\]
While any positive number, (except one), can be the base of a logarithm, the bases [latex]10[/latex] and [latex]e[/latex] are the most common. It is important to understand various notations. Usually [latex]\log[/latex] means [latex]\log_{10}[/latex] but there are exceptions. The logarithm to base [latex]e[/latex] is called the natural logarithm and [latex]\ln[/latex] means [latex]\log_e[/latex].
Logarithms and the calculator
Most calculators have a LOG button for [latex]\log_{10}[/latex] and an LN button for [latex]\log_e[/latex].
To find the power of [latex]10[/latex] that produces [latex]50[/latex] we use the LOG button on the calculator: [latex]\log_{10}50\approx1.7[/latex] and [latex]10^{1.7}\approx50[/latex].
To find the power of [latex]e[/latex] that produces [latex]50[/latex] we use the LN button on the calculator: [latex]\log_{e}50=\ln50\approx3.9[/latex] and [latex]e^{3.9}\approx50[/latex].
You will usually need a calculator to calculate logarithms. But if the bases and the number are simply powers, you can do some in your head as shown in the examples below.
Here are some exercises to try.
Exercises: Calculating logs using a calculator
Determine the following with your calculator.
1. [latex]\log_{10} \left( 3 \right)[/latex]
2. [latex]\log_{10} \left( 25 \right)[/latex]
3. [latex]\log_{10} \left( 130 \right)[/latex]
4. [latex]\ln \left( 1.5 \right)[/latex]
5. [latex]\ln \left( 8 \right)[/latex]
Answers
[latex]\begin{array}{lllllll} \ \qquad 1.\ 0.4771 & 2.\ 1.3979 & 3.\ 2.1139 & 4.\ 0.4055 & 5.\ 2.0794 \end{array}[/latex]
In special cases, you do not need a calculator as shown in the following examples.
Examples: Calculating logs without a calculator
1. For powers of 10, it is easy to calculate the logarithm in base 10. For example
[latex]\quad \log_{10} \left( 10 \right) = \log_{10} \left( 10^1 \right) = 1[/latex]
[latex]\quad \log_{10} \left( 100 \right) = \log_{10} \left( 100^2 \right) = 2[/latex]
[latex]\quad \log_{10} \left( 1000 \right) = \log_{10} \left( 10^3 \right) = 3[/latex].
2. Note that logarithms can be negative. For example:
[latex]\quad \log_{10} \left( 0.10 \right) = \log_{10} \left( 1/10 \right) = \log_{10} \left(10^{-1} \right)=- 1[/latex]
[latex]\quad \log_{10} \left( 0.0001 \right) = \log_{10} \left( 1/10000 \right) = \log_{10} \left(10^{-4} \right)=- 4[/latex].
3. When the number is the base raised to an integer, you can determine the logarithm without a calculator. For example:
[latex]\quad \log_5 \left( 125 \right) = \log_5 \left( 5^3 \right) = 3[/latex]
[latex]\quad \log_7 \left( 49 \right) = \log_7 \left( 7^2 \right) = 2[/latex]
[latex]\quad \log_3 \left( 1/81\right) = \log_3 \left( 3^{-4} \right) = -4[/latex].
Here are some exercises to try.
Exercises: Calculating logs without a calculator
Evaluate without using a calculator
1. [latex]\log_{6}36[/latex]
2. [latex]\log_{10}\sqrt{10}[/latex]
3. [latex]\log_{5}1[/latex]
4. [latex]\log_{10}100000[/latex]
5. [latex]\log_{5}5[/latex]
Answers
[latex]\begin{array}{lllllll} \quad \mathbf{1.}\ 2 & \mathbf{2.}\ 1/2 & \mathbf{3.}\ 0 & \mathbf{4.}\ 5 & \mathbf{5.}\ 1 \end{array}[/latex]
Change of base
While the most common bases in mathematics, science and engineering are [latex]e[/latex] and [latex]10[/latex], some fields use other bases. In computer science and software engineering the use of base [latex]2[/latex] is common. A conventional calculator can be used to determine a logarithm with a different base [latex]a[/latex] using the formula:
[latex]\begin{align*} \log _{a}\left(x \right) &= \frac{\log _{c} \left(x \right)}{\log _{c}\left(a \right)} \end{align*}[/latex]
where [latex]c[/latex] is any base. On conventional calculators, [latex]c[/latex] will be [latex]10[/latex] or [latex]e[/latex].
For example, if you want to determine [latex]\log_{2}\left(5 \right)[/latex], you can use [latex]\log_{10}[/latex] to get,
[latex]\begin{align*} \log_{2}\left(5 \right) &= \frac{\log _{10} \left(5 \right)}{\log _{10}\left(2 \right)}\\ &\approx\frac{0.6990}{0.3010}\\ &\approx 2.3219. \end{align*}[/latex]
Alternatively, using base [latex]e[/latex],
[latex]\begin{align*} \log_{2}\left(5 \right) &= \frac{\log _{e} \left(5 \right)}{\log _{e}\left(2 \right)}\\ &\approx \frac{1.6094}{0.6931}\\ &\approx 2.3219 \end{align*}[/latex]
as before.
Three important properties of logarithms
1. For [latex]a>0[/latex],
[latex]\begin{align*} \log_{a}1=0. \end{align*}[/latex]
That is, for any [latex]a>0[/latex], the log of [latex]1[/latex] is zero.
2. For [latex]a>0[/latex],
[latex]\begin{align*} \log_{a}a=1. \end{align*}[/latex]
3. If [latex]\log_{a}m=\log_{a}p[/latex] then [latex]m=p[/latex] .
Logarithm laws
Corresponding to the three index laws, there are three laws of logarithms to help in manipulating logarithms. If [latex]m,n>0[/latex] and [latex]a>0,\ a\neq1[/latex] then
1. First Logarithm Law
[latex]\begin{align*} \log_{a}(mn) & =\log_{a}m+\log_{a}n \end{align*}[/latex]
2. Second Logarithm Law
[latex]\begin{align*} \log_{a}\frac{m}{n} & =\log_{a}m-\log_{a}n. \end{align*}[/latex]
3. Third Logarithm Law
[latex]\begin{align*} \log_{a}m^{p} & =p\log_{a}m. \end{align*}[/latex]
The laws can be used to simplify and evaluate logarithmic expressions.
Examples: The log laws
1. Use of first law:
[latex]\begin{align*} \log_{a}15=\log_{a}(5\times3)&=\log_{a}5+\log_{a}3. \end{align*}[/latex]
2. Use of second law:
[latex]\begin{align*} \log_{a}\frac{12}{5}=\log_{a}12-\log_{a}5. \end{align*}[/latex]
3. Use of third law:
[latex]\begin{align*} \log_{a}\left(9^{2}\right)=2\log_{a}9=2\log_{a}3^{2}=4\log_{a}3. \end{align*}[/latex]
4. Simplifying using the first law:
[latex]\begin{align*} \log_{10}6+\log_{10}2=\log_{10}(6\times2)=\log_{10}12. \end{align*}[/latex]
5. Simplify:
[latex]\begin{align*} \log_{3}6a+\log_{3}b-\log_{3}2a. \end{align*}[/latex]
Solution:
[latex]\begin{align*} \log_{3}6a+\log_{3}b-\log_{3}2a & =\log_{3}\left(6ab \right)-\log_{3} \left( 2a\right)\quad \text{(using first law)}\\ &=\log_{3}\frac{\left(6ab \right)}{ 2a}\quad \text{(using second law)}\\ &=\log_{3}3b. \end{align*}[/latex]
6. Simplify:
[latex]\begin{align*} \frac{1}{2}\log_{10}36-\log_{10}15+2\log_{10}5. \end{align*}[/latex]
Solution:
[latex]\begin{align*} \frac{1}{2}\log_{10}36-\log_{10}15+2\log_{10}5 & =\log_{10}36^{\frac{1}{2}}-\log_{10}15+\log_{10}5^{2} \\ &\qquad \text{(using third law twice)}\\ & =\log_{10}6-\log_{10}15+\log_{10}25\\ & =\log_{10}\frac{6\times25}{15} \quad \text{(using first and second laws)}\\ & =\log_{10}10\\ & =1. \end{align*}[/latex]
Here are some exercises to try.
Exercises: Log laws
Simplify the following
1. [latex]\log_{4}8+\log_{4}3-\log_{4}2[/latex]
2. [latex]\dfrac{1}{2}\log_{10}25-\log_{10}4+2\log_{10}3 [/latex]
3. [latex]\log_{e}e^{2}+2\log_{e}3-\log_{e}18 [/latex]
4. [latex]\log_{a}4+2\log_{a}3-2\log_{a}6[/latex]
5. [latex]\dfrac{1}{2}\log_{10}a^{2}+3\log_{10}b-\log_{10}3ab^{2}[/latex]
Answers
[latex]\begin{array}{lllllll} \quad \mathbf{1.}\ \log_{4}12 & \mathbf{2.}\ \log_{10}\left(\dfrac{45}{4} \right) & \mathbf{3.}\ 2-\log_{e} 2 & \mathbf{4.}\ 0 & \mathbf{5.}\ \log_{10}\left(\dfrac{b}{3} \right). \end{array}[/latex]
The logarithmic function
This section briefly considers the function [latex]y=\log_{a}x[/latex] which is the log of [latex]x[/latex] to base [latex]a[/latex] where [latex]a>0[/latex] and [latex]a \neq 1[/latex]. The effect of changing the base on the graph of [latex]y=\log_{a}x[/latex] may be seen in the link: Interactive graph of [latex]y=\log_{a}\left(x\right)[/latex]
By changing the value of [latex]a[/latex] using the slider, you can see how the base changes the graph. Note that regardless of base, the value of [latex]\log_{a} \left(1\right)=0[/latex].
Usually, the base will be [latex]10[/latex] or [latex]e[/latex]. Logarithmic functions to the base [latex]10[/latex] are written as [latex]\log\left(x \right)[/latex] or [latex]\log_{10} \left(x \right)[/latex] whereas logarithms to the base [latex]e[/latex] are written called natural logarithms and denoted by [latex]\ln \left(x \right)[/latex] or [latex]\log_e \left(x \right)[/latex]. Natural logarithms are very common in the physical sciences and mathematics.
Note that sometimes the natural log function is denoted as [latex]\log\left(x \right)[/latex].
The general natural log function is of the form:
[latex]\begin{align*} f\left( x \right) = a\ln \left(cx \right)+d, \end{align*}[/latex]
where [latex]a[/latex], [latex]b[/latex], [latex]c[/latex] and [latex]d[/latex] are real constants.
Click here to see an interactive graph of the natural log function. Use the sliders to change the values of the constants. You will find that [latex]a[/latex] stretches the graph parallel to the [latex]y[/latex]-axis, [latex]b[/latex] increases the slope of the graph [latex]c[/latex] shifts the asymptote by moving the graph horizontally and [latex]d[/latex] shifts the graph vertically up or down.
The exponential function
The exponential function is denoted by [latex]b^x[/latex] where [latex]b[/latex] is called the base.
If the base is [latex]10[/latex] the exponential function is denoted as [latex]f\left(x \right) = 10^x[/latex].
If the base is [latex]e[/latex], the exponential function is denoted as [latex]e^x[/latex] or [latex]\exp \left( x \right)[/latex]. The latter form is used when you are exponentiating another function or expression. For example, we would write
[latex]\begin{align*} e^{\int \frac{1}{2} \sin^2 \left(x \right) dx } \end{align*}[/latex]
as
[latex]\begin{align*} \exp\left(\int \frac{1}{2}\sin^2 \left(x \right) \right). \end{align*}[/latex]
The general exponential function, in base [latex]e[/latex], is
[latex]\begin{align*} y&= ae^{bx+c}+d, \end{align*}[/latex]
where [latex]a[/latex], [latex]b[/latex], [latex]c[/latex] and [latex]d[/latex] are real constants. Click here to see an interactive graph of the natural exponential function. Use the sliders to change the values of the constants. You will find that [latex]a[/latex] stretches the graph parallel to the [latex]y[/latex]-axis, [latex]b[/latex] increases the slope of the graph, [latex]c[/latex] moves the graph horizontally, and [latex]d[/latex] shifts the graph vertically up or down.
Inverses of logarithmic and exponential functions
The logarithmic function is the inverse of exponentiation, and exponentiation is the inverse of the logarithm. The following figure gives the graphs of [latex]y=e^x[/latex] in blue and [latex]y=\ln\left(x\right)[/latex] in red. Also shown is the graph of [latex]y=x[/latex] in dashed black. As expected, the graph of [latex]y=\ln\left(x \right)[/latex] is the reflection of [latex]y=e^x[/latex] about the line [latex]y=x[/latex].
Logarithmic equations
The laws of logarithms can be used to solve equations involving logarithms or in which the variable is a power. An example of an equation involving logs is
[latex]\begin{align*} \log_{10} x- \log_{10}3=2 \end{align*}[/latex]
and may be solved as follows. Using the second law, the equation may be written
[latex]\begin{align*} \log_{10}\frac{x}{3} =2. \end{align*}[/latex]
Remembering that exponentiation is the inverse operation of taking logs,
[latex]\begin{align*} 10^{\log_{10}\frac{x}{3} }=10^2. \end{align*}[/latex]
Now using the definition of the logarithm,
[latex]\begin{align*} \frac{x}{3} &=100\\ x &=\frac{100}{3}. \end{align*}[/latex]
An example of an exponential equation is
[latex]\begin{align*} 2^{2x+1} &=5^{2-x}.\\ \end{align*}[/latex]
This may be solved by taking the logarithm of both sides and solving for [latex]x[/latex] as follows.
[latex]\begin{align*} \log_{10} \left(2^{2x+1}\right) &=\log_{10} \left(5^{2-x}\right)\\ \left(2x+1\right) \log_{10} \left(2\right)&= \left(2-x\right)\log_{10} \left(5\right)\quad\text{(using the third law)}\\ \left(2x+1\right)0.301 &=\left(2-x\right)0.699\quad\text{(using a calculator)}\\ 0.602x+0.301 &= 1.398 -0.699x\\ 1.301x &= 1.097\\ x &= 0.843. \end{align*}[/latex]
We use these techniques in the examples below.
Examples: Solving log and exponential equations
1. Solve for [latex]x[/latex] if [latex]\log_{10}(2x+1)=\log_{10}3[/latex]
Solution:
[latex]\begin{align*} \log_{10}(2x+1)= & \log_{10}3\\ \end{align*}[/latex]
Exponentiating each side of the equation
[latex]\begin{align*} 10^{\log_{10}(2x+1)}= & 10^{\log_{10}3}\\ \implies 2x+1 & =3\quad \text{(using definition of the logarithm)}\\ \implies 2x & =2\\ \implies x &=1. \end{align*}[/latex]
2. Solve for [latex]x[/latex] if [latex]2^{x-3}=10[/latex]
Solution:
[latex]\begin{align*}2^{x-3} & =10\\ \log_{10}2^{x-3} & =\log_{10}10\ \text{(take logs of both sides)}\\ (x-3)\log_{10}2 & =\log_{10}10\\ (x-3)(0.301) & =1\ (\textrm{use a calculator to find}\log_{10}2)\\ x-3 & =\frac{1}{0.301}\\ x & =3+\frac{1}{0.301}\\ x & =6.32 \end{align*}[/latex]
3. The number of bacteria present in a sample is given by [latex]N=800e^{0.2t}[/latex], where [latex]t[/latex] is the time in seconds. Find a) the initial number of bacteria and b) the time taken for the bacteria count to reach 10000.
Solution a)
The initial number of bacteria is the number when [latex]t=0[/latex]:
[latex]\begin{align*} N & =800e^{0.2t}\\ & =800e^{0.2\times0}\\ & =800e^{0}\\ & =800. \end{align*}[/latex]
Solution b)
To find [latex]t[/latex] when [latex]N=10000[/latex], set N=10000 in the equation:
[latex]\begin{align*} N & =800e^{0.2t}\\ 10000 & =800e^{0.2t}\\ \frac{10000}{800} & =e^{0.2t}\\ 12.5 & =e^{0.2t}\\ \ln12.5 & =\ln e^{0.2t} \quad\textrm{(use ln when the base is }e)\\ \ln12.5 & =0.2t\ln e \quad\text{(using the third law)}\\ \ln12.5 & =0.2t\ \quad \text{(remember }\ln\left( e\right) =\log_{e}e=1)\\ t & =\frac{\ln12.5}{0.2}\\ t & =12.7. \end{align*}[/latex]
It takes [latex]12.7[/latex] sec for the number of bacteria to reach [latex]10000[/latex].
Here are some exercises to try.
Exercises: Solving log and exponential equations
Solve the following for [latex]x[/latex]:
[latex]\begin{align*} \quad& \mathbf{1.} \ \log_{10}x=\log_{10}4-\log_{10}2\\ \quad&\mathbf{2.}\ \log_{2}2x-2\log_{2}3=\log_{2}6\\ \quad&\mathbf{3.}\ 10^{x+1}=7 \end{align*}[/latex]
Answers
[latex]\begin{array}{lll} \quad \mathbf{1.}\ x=2 & \mathbf{2.}\ x=27 & \mathbf{3.}\ x=-0.155\end{array}[/latex]
Key takeaways
1. The logarithm [latex]a[/latex], in a particular base [latex]b[/latex], of a number [latex]x[/latex] is the index (or power) to which the base must be raised to produce the number. That is
[latex]\begin{align*} x&=b^a\\ &=b^{\log_b\left(x \right)}. \end{align*}[/latex]
2. For any [latex]a\gt0[/latex], [latex]\log_a\left(a\right)=1[/latex].
3. For any [latex]a\gt0[/latex], [latex]\log_a\left(1 \right)=0[/latex].
3. Log rules:
[latex]\begin{align*} \log\left( mn \right)&=\log\left( m \right) + \log\left( n \right)\\ \log\left( \frac{m}{n} \right)&=\log\left( m \right) - \log\left( n \right)\\ \log\left( x^n \right)&=n\log\left(x \right). \end{align*}[/latex]
4. The natural log, [latex]\log_e[/latex] and the exponential function [latex]e^x[/latex]or [latex]\exp\left(x \right)[/latex] are inverse functions:
[latex]\begin{align*} \log_e\left( e^x \right)&=x=e^{\log_{e} x}=\exp\left( \log_e x\right). \end{align*}[/latex]
