3.3 Logarithms

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Contents:

  • Logarithms
  • Fundamental Laws of Logarithms

Learning outcomes:

After this section, you will have learned:

  • The concepts of base and exponent.
  • The meaning of the notation \displaystyle \ln, \displaystyle \lg, \displaystyle \log and \displaystyle \log_{a}.
  • To calculate simple logarithmic expressions using the definition of a logarithm.
  • That logarithms are only defined for positive numbers.
  • The value of the number \displaystyle e.
  • To use the laws of logarithms to simplify logarithmic expressions.
  • To know when the laws of logarithms are valid.
  • To express a logarithm in terms of a logarithm with a different base.

Logarithms to the base 10

We often use powers with base \displaystyle 10 to represent large and small numbers, for example

\displaystyle \begin{align*} 
   10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\
   10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0\textrm{.}01\,\mbox{.}
 \end{align*}

If we only consider the exponents, we can informally state that

"the exponent of 1000 is 3",

or

"the exponent of 0.01 is -2".

This is how logarithms are defined. We formalise this as follows:

"The logarithm of 1000 is 3",

which is written as \displaystyle \lg 1000 = 3, and

"The logarithm of 0.01 is -2",

which is written as \displaystyle \lg 0\textrm{.}01 = -2.

More generally, we say:

The logarithm of a number \displaystyle y is denoted by \displaystyle \lg y and is the real number \displaystyle x in the blue box which satisfies the equality
\displaystyle 10^{\ \bbox[#AAEEFF,2pt]{\,x\,}} = y\,\mbox{.}

Note that \displaystyle y must be a positive number for the logarithm \displaystyle \lg y to be defined, since there is no power of 10 that evaluates to a negative number, or for that matter zero .

Example 1

  1. \displaystyle \lg 100000 = 5\quad because \displaystyle 10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} = 100\,000.
  2. \displaystyle \lg 0\textrm{.}0001 = -4\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} = 0\textrm{.}0001.
  3. \displaystyle \lg \sqrt{10} = \frac{1}{2}\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}.
  4. \displaystyle \lg 1 = 0\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
  5. \displaystyle \lg 10^{78} = 78\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} = 10^{78}.
  6. \displaystyle \lg 50 \approx 1\textrm{.}699\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\textrm{.}699\,}} \approx 50.
  7. \displaystyle \lg (-10) does not exist because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}} can never be -10 regardless of how \displaystyle a is chosen.

In the penultimate example, one can easily understand that \displaystyle \lg 50 must lie somewhere between 1 and 2 since \displaystyle 10^1 < 50 < 10^2. However, in practice to obtain a more precise value of the irrational number \displaystyle \lg 50 = 1\textrm{.}69897\ldots one needs a calculator (or table).

Example 2

  1. \displaystyle 10^{\textstyle\,\lg 100} = 100
  2. \displaystyle 10^{\textstyle\,\lg a} = a
  3. \displaystyle 10^{\textstyle\,\lg 50} = 50


Different bases

We can define logarithms with respect to a base other than 10. We must clearly indicate which number is used as the base for the logarithm. If we use a base such as 2 we write \displaystyle \log_{\,2} to mean a logarithm with base 2.

Example 3

  1. \displaystyle \log_{\,2} 8 = 3\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8.
  2. \displaystyle \log_{\,2} 2 = 1\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2.
  3. \displaystyle \log_{\,2} 1024 = 10\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024.
  4. \displaystyle \log_{\,2}\frac{1}{4} = -2\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2} = \frac{1}{4}.

We deal with logarithms which have other bases in the same way.

Example 4

  1. \displaystyle \log_{\,3} 9 = 2\quad because \displaystyle 3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9.
  2. \displaystyle \log_{\,5} 125 = 3\quad because \displaystyle 5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125.
  3. \displaystyle \log_{\,4} \frac{1}{16} = -2\quad because \displaystyle 4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2} = \frac{1}{16}.
  4. \displaystyle \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad as \displaystyle b^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-1/2\,}} = \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}} (if \displaystyle b>0 and \displaystyle b\not=1).

If the base 10 is used, one rarely writes \displaystyle \log_{\,10}, but as we have previously seen we use the notation \displaystyle \lg , or simply \displaystyle \log. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base \displaystyle 1.


The natural logarithms

In practice there are two bases that are commonly used for logarithms, 10 and the number \displaystyle e \displaystyle ({}\approx 2\textrm{.}71828 \ldots\,). Logarithms using the base e are called natural logarithms and we use the notation \displaystyle \ln instead of \displaystyle \log_{\,e}.

Example 5

  1. \displaystyle \ln 10 \approx 2{\textrm{.}}3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{\textrm{.}}3\,}} \approx 10.
  2. \displaystyle \ln e = 1\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e.
  3. \displaystyle \ln\frac{1}{e^3} = -3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}} = \frac{1}{e^3}.
  4. \displaystyle \ln 1 = 0\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
  5. If \displaystyle y= e^{\,a} then \displaystyle a = \ln y.
  6. \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5
  7. \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x

Most advanced calculators have buttons for 10-logarithms and natural logarithms.


Laws of Logarithms

Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition is a much faster calculation than multiplication).

Example 6

Calculate \displaystyle \,35\cdot 54.

If we know that \displaystyle 35 \approx 10^{\,1\textrm{.}5441} and \displaystyle 54 \approx 10^{\,1\textrm{.}7324} (i.e. \displaystyle \lg 35 \approx 1\textrm{.}5441 and \displaystyle \lg 54 \approx 1\textrm{.}7324) then we can calculate that

\displaystyle 
 35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324}
             = 10^{\,1\textrm{.}5441 + 1\textrm{.}7324}
             = 10^{\,3\textrm{.}2765}.

Since \displaystyle 10^{\,3\textrm{.}2765} \approx 1890 (i.e. \displaystyle \lg 1890 \approx 3\textrm{.}2765), we have thus managed to calculate the product

\displaystyle 35 \cdot 54 = 1890

just by adding together the exponents \displaystyle 1\textrm{.}5441 and \displaystyle 1\textrm{.}7324.

In the above example we have used a logarithmic law which states that

\displaystyle \log (ab) = \log a + \log b,

for all \displaystyle a,b>0. We can see that this is true by using the laws of exponents. Indeed,

\displaystyle 
 a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b}
          = 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}

and on the other hand,

\displaystyle 
 a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}

so that we can see \displaystyle \log (ab) = \log a + \log b. By exploiting the laws of exponents in this way we can obtain the corresponding laws of logarithms:

\displaystyle \begin{align*} 
   \log(ab) &= \log a + \log b,\\[4pt]
   \log\frac{a}{b} &= \log a - \log b,\\[4pt]
   \log a^b &= b\cdot \log a\,\mbox{,}\\
 \end{align*}

for \displaystyle a,b>0. These laws hold regardless of the base we are working with.

Example 7

  1. \displaystyle \lg 4 + \lg 7 = \lg(4 \cdot 7) = \lg 28
  2. \displaystyle \lg 6 - \lg 3 = \lg\frac{6}{3} = \lg 2
  3. \displaystyle 2 \cdot \lg 5 = \lg 5^2 = \lg 25
  4. \displaystyle \lg 200 = \lg(2 \cdot 100) = \lg 2 + \lg 100 = \lg 2 + 2

Example 8

  1. \displaystyle \lg 9 + \lg 1000 - \lg 3 + \lg 0\textrm{.}001 = \lg 9 + 3 - \lg 3 - 3 = \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3
  2. \displaystyle \ln\frac{1}{e} + \ln \sqrt{e} = \ln\left(\frac{1}{e} \cdot \sqrt{e}\,\right) = \ln\left( \frac{1}{(\sqrt{e}\,)^2} \cdot \sqrt{e}\,\right) = \ln\frac{1}{\sqrt{e}}
    \displaystyle \phantom{\ln\frac{1}{e} + \ln \sqrt{e}}{} = \ln e^{-1/2} = -\frac{1}{2} \cdot \ln e =-\frac{1}{2} \cdot 1 = -\frac{1}{2}\vphantom{\biggl(}
  3. \displaystyle \log_2 36 - \frac{1}{2} \log_2 81 = \log_2 (6 \cdot 6) - \frac{1}{2} \log_2 (9 \cdot 9)
    \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 (2\cdot 2 \cdot 3 \cdot 3) - \frac{1}{2} \log_2 (3 \cdot 3 \cdot 3 \cdot 3)
    \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 (2^2 \cdot 3^2) - \frac{1}{2} \log_2 (3^4)\vphantom{\Bigl(}
    \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 2^2 + \log_2 3^2 - \frac{1}{2} \log_2 3^4
    \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = 2 \log_2 2 + 2 \log_2 3 - \frac{1}{2} \cdot 4 \log_2 3
    \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = 2\cdot 1 + 2 \log_2 3 - 2 \log_2 3 = 2\vphantom{\Bigl(}
  4. \displaystyle \lg a^3 - 2 \lg a + \lg\frac{1}{a} = 3 \lg a - 2 \lg a + \lg a^{-1}
    \displaystyle \phantom{\lg a^3 - 2 \lg a + \lg\frac{1}{a}}{} = (3-2)\lg a + (-1) \lg a = \lg a - \lg a = 0


Changing the base

It is sometimes a good idea to express a logarithm as a logarithm with respect to another base.

Example 9

  1. Express \displaystyle \lg 5 as a natural logarithm.

    By definition, \displaystyle \lg 5 is a number that satisfies the equality
    \displaystyle 10^{\lg 5} = 5\,\mbox{.}

    Taking the natural logarithm ( \displaystyle \ln) of both sides yields

    \displaystyle \ln 10^{\lg 5} = \ln 5\,\mbox{.}

    With the help of the logarithm law \displaystyle \ln a^b = b \ln a, the left-hand side can be written as \displaystyle \lg 5 \cdot \ln 10 and the equality becomes

    \displaystyle \lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}

    Now divide both sides by \displaystyle \ln 10 to get the answer

    \displaystyle 
     \lg 5 = \frac{\ln 5}{\ln 10}
 \qquad (\approx 0\textrm{.}699\,,
 \quad\text{dvs.}\ 10^{0\textrm{.}699} \approx 5)\,\mbox{.}

  2. Express the 2-logarithm of 100 as a 10-logarithm lg.

    Using the definition of a logarithm one has that \displaystyle \log_2 100 formally satisfies
    \displaystyle 2^{\log_{\scriptstyle 2} 100} = 100.

    Taking the 10-logarithm of both sides, we get

    \displaystyle 
     \lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}

    Since \displaystyle \lg a^b = b \lg a we get \displaystyle \lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2, and moreover the right-hand side can be simplified to \displaystyle \lg 100 = 2. Thus we see that

    \displaystyle 
     \log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}

    Finally, dividing by \displaystyle \lg 2 gives

    \displaystyle 
     \log_{\scriptstyle 2} 100 = \frac{2}{\lg 2}
 \qquad ({}\approx 6\textrm{.}64\,,
 \quad\text{that is}\ 2^{6\textrm{.}64}\approx 100 )\,\mbox{.}

The general formula for changing from one base \displaystyle a to another base \displaystyle b is

\displaystyle 
 \log_{\scriptstyle\,a} x
  = \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a}
  \,\mbox{,}

and can be derived in the same way. We can also change the base of a power using logarithms. For instance, if we want to write \displaystyle 2^5 using the base 10 first write 2 as a power with the base 10:

\displaystyle 2 = 10^{\lg 2}.

Then, using one of the laws of exponents,

\displaystyle 
 2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2}
 \quad ({}\approx 10^{1\textrm{.}505}\,)\,\mbox{.}

Example 10

  1. Write \displaystyle 10^x using the base e.

    First, we write 10 as a power of e,
    \displaystyle 10 = e^{\ln 10}.

    Using the the laws of exponents we can then see that

    \displaystyle 
     10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10}
      \approx e^{2\textrm{.}3 x}\,\mbox{.}
  2. Write \displaystyle e^{\,a} using the base 10.

    The number \displaystyle e can be written as \displaystyle e=10^{\lg e} and therefore
    \displaystyle 
     e^a = (10^{\lg e})^a
     = 10^{\,a \cdot \lg e}
     \approx 10^{\,0\textrm{.}434a}\,\mbox{.}


Exercises

Study advice

The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that...

You may need to spend some time studying logarithms.

Logarithms are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.


Reviews

For those of you who want to deepen your understanding or need more detailed explanations consider the following references:

Learn more about logarithms on Wikipedia

Learn more about the number e in The MacTutor History of Mathematics archive


Useful web sites

Experiment with logarithms and powers

Play logarithm Memory

Help the frog to jump onto his water-lily leaf in the "log" game

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