3.3 Logarithmen
Aus Online Mathematik Brückenkurs 1
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We often use powers with base <math>10</math> to represent large and small numbers, for example, | We often use powers with base <math>10</math> to represent large and small numbers, for example, | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*} |
10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\ | 10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\ | ||
10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0\textrm{.}01\,\mbox{.} | 10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0\textrm{.}01\,\mbox{.} | ||
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:::The logarithm of a number <math>y</math> is designated by <math>\lg y</math> and is the exponent in the blue box which satisfies the equality | :::The logarithm of a number <math>y</math> is designated by <math>\lg y</math> and is the exponent in the blue box which satisfies the equality | ||
| - | {{ | + | {{Displayed math||<math>10^{\ \bbox[#AAEEFF,2pt]{\,\phantom{a}\,}} = y\,\mbox{.} </math>}} |
Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number or for that matter zero . | Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number or for that matter zero . | ||
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If we know that <math>35 \approx 10^{\,1\textrm{.}5441}</math> and <math>54 \approx 10^{\,1\textrm{.}7324}</math> (i.e. <math>\lg 35 \approx 1\textrm{.}5441</math> and <math>\lg 54 \approx 1\textrm{.}7324</math>) then we can calculate that | If we know that <math>35 \approx 10^{\,1\textrm{.}5441}</math> and <math>54 \approx 10^{\,1\textrm{.}7324}</math> (i.e. <math>\lg 35 \approx 1\textrm{.}5441</math> and <math>\lg 54 \approx 1\textrm{.}7324</math>) then we can calculate that | ||
| - | {{ | + | {{Displayed math||<math> |
35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324} | 35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324} | ||
= 10^{\,1\textrm{.}5441 + 1\textrm{.}7324} | = 10^{\,1\textrm{.}5441 + 1\textrm{.}7324} | ||
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and we then know that <math>10^{\,3\textrm{.}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3\textrm{.}2765</math>) thus we have managed to calculate the product | and we then know that <math>10^{\,3\textrm{.}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3\textrm{.}2765</math>) thus we have managed to calculate the product | ||
| - | {{ | + | {{Displayed math||<math>35 \cdot 54 = 1890</math>}} |
and this just by adding together exponents <math>1\textrm{.}5441</math> and <math>1\textrm{.}7324</math>. | and this just by adding together exponents <math>1\textrm{.}5441</math> and <math>1\textrm{.}7324</math>. | ||
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This is an example of a logarithmic law which says that | This is an example of a logarithmic law which says that | ||
| - | {{ | + | {{Displayed math||<math>\log (ab) = \log a + \log b</math>}} |
This stems from the fact that on the one hand, | This stems from the fact that on the one hand, | ||
| - | {{ | + | {{Displayed math||<math> |
a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b} | a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b} | ||
= \left\{ \mbox{laws of exponents} \right\} | = \left\{ \mbox{laws of exponents} \right\} | ||
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and on the other hand, | and on the other hand, | ||
| - | {{ | + | {{Displayed math||<math> |
a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}} | a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}} | ||
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<div class="regel"> | <div class="regel"> | ||
| - | {{ | + | {{Displayed math||<math>\begin{align*} |
\log(ab) &= \log a + \log b,\\[4pt] | \log(ab) &= \log a + \log b,\\[4pt] | ||
\log\frac{a}{b} &= \log a - \log b,\\[4pt] | \log\frac{a}{b} &= \log a - \log b,\\[4pt] | ||
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<br> | <br> | ||
By definition, the <math>\lg 5</math> is a number that satisfies the equality | By definition, the <math>\lg 5</math> is a number that satisfies the equality | ||
| - | {{ | + | {{Displayed math||<math>10^{\lg 5} = 5\,\mbox{.}</math>}} |
Take the natural logarithm ( ln ) of both sides. | Take the natural logarithm ( ln ) of both sides. | ||
| - | {{ | + | {{Displayed math||<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}} |
With the help of the logarithm law <math>\ln a^b = b \ln a</math> the left-hand side can be written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes | With the help of the logarithm law <math>\ln a^b = b \ln a</math> the left-hand side can be written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes | ||
| - | {{ | + | {{Displayed math||<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}} |
Now divide both sides by <math>\ln 10</math> giving the answer | Now divide both sides by <math>\ln 10</math> giving the answer | ||
| - | {{ | + | {{Displayed math||<math> |
\lg 5 = \frac{\ln 5}{\ln 10} | \lg 5 = \frac{\ln 5}{\ln 10} | ||
\qquad (\approx 0\textrm{.}699\,, | \qquad (\approx 0\textrm{.}699\,, | ||
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<br> | <br> | ||
Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies | Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies | ||
| - | {{ | + | {{Displayed math||<math>2^{\log_{\scriptstyle 2} 100} = 100</math>}} |
and taking the 10-logarithm (lg) of both sides, one gets | and taking the 10-logarithm (lg) of both sides, one gets | ||
| - | {{ | + | {{Displayed math||<math> |
\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}} | \lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}} | ||
Since <math>\lg a^b = b \lg a</math> one gets <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math> and the right-hand side can be simplified to <math>\lg 100 = 2</math>. This gives the equality | Since <math>\lg a^b = b \lg a</math> one gets <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math> and the right-hand side can be simplified to <math>\lg 100 = 2</math>. This gives the equality | ||
| - | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}} | \log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}} | ||
Finally, dividing by <math>\lg 2</math> gives that | Finally, dividing by <math>\lg 2</math> gives that | ||
| - | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2} | \log_{\scriptstyle 2} 100 = \frac{2}{\lg 2} | ||
\qquad ({}\approx 6\textrm{.}64\,, | \qquad ({}\approx 6\textrm{.}64\,, | ||
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The general formula for changing from one base <math>a</math> to another base <math>b</math> can be derived in the same way | The general formula for changing from one base <math>a</math> to another base <math>b</math> can be derived in the same way | ||
| - | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle\,a} x | \log_{\scriptstyle\,a} x | ||
= \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a} | = \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a} | ||
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If one wants to change the base of a power, one can do this by using logarithms. For instance, if we want to write <math> 2^5 </math> using the base 10 one first writes 2 as a power with the base 10; | If one wants to change the base of a power, one can do this by using logarithms. For instance, if we want to write <math> 2^5 </math> using the base 10 one first writes 2 as a power with the base 10; | ||
| - | {{ | + | {{Displayed math||<math>2 = 10^{\lg 2}</math>}} |
and then using one of the laws of exponents | and then using one of the laws of exponents | ||
| - | {{ | + | {{Displayed math||<math> |
2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2} | 2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2} | ||
\quad ({}\approx 10^{1\textrm{.}505}\,)\,\mbox{.}</math>}} | \quad ({}\approx 10^{1\textrm{.}505}\,)\,\mbox{.}</math>}} | ||
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<br> | <br> | ||
First, we write 10 as a power of ''e'', | First, we write 10 as a power of ''e'', | ||
| - | {{ | + | {{Displayed math||<math>10 = e^{\ln 10}</math>}} |
and then use the laws of exponents | and then use the laws of exponents | ||
| - | {{ | + | {{Displayed math||<math> |
10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} | 10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} | ||
\approx e^{2\textrm{.}3 x}\,\mbox{.}</math>}}</li> | \approx e^{2\textrm{.}3 x}\,\mbox{.}</math>}}</li> | ||
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<br> | <br> | ||
The number <math>e</math> can be written as <math>e=10^{\lg e}</math> and therefore | The number <math>e</math> can be written as <math>e=10^{\lg e}</math> and therefore | ||
| - | {{ | + | {{Displayed math||<math> |
e^a = (10^{\lg e})^a | e^a = (10^{\lg e})^a | ||
= 10^{\,a \cdot \lg e} | = 10^{\,a \cdot \lg e} | ||
Version vom 13:55, 10. Sep. 2008
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 meaning 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,
If one only considers the exponents one can state that
- "the exponent for 1000 is 3", or
- "the exponent for 0.01 is -2".
This is how logarithms are defined. One formalises this as follows:
- "The logarithm of 1000 is 3", which is written as \displaystyle \lg 1000 = 3,
- "The logarithm of 0.01 is -2", which is written as \displaystyle \lg 0\textrm{.}01 = -2.
More generally, one says:
- The logarithm of a number \displaystyle y is designated by \displaystyle \lg y and is the exponent in the blue box which satisfies the equality
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
- \displaystyle \lg 100000 = 5\quad because \displaystyle 10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} = 100\,000.
- \displaystyle \lg 0\textrm{.}0001 = -4\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} = 0\textrm{.}0001.
- \displaystyle \lg \sqrt{10} = \frac{1}{2}\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}.
- \displaystyle \lg 1 = 0\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
- \displaystyle \lg 10^{78} = 78\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} = 10^{78}.
- \displaystyle \lg 50 \approx 1\textrm{.}699\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\textrm{.}699\,}} \approx 50.
- \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, but to obtain a more precise value of the irrational number \displaystyle \lg 50 = 1\textrm{.}69897\ldots one needs in practice, a calculator (or table.)
Example 2
- \displaystyle 10^{\textstyle\,\lg 100} = 100
- \displaystyle 10^{\textstyle\,\lg a} = a
- \displaystyle 10^{\textstyle\,\lg 50} = 50
Different bases
One can imagine logarithms, which use a base other than 10 (except 1!). One must clearly indicate which number is used as a base for a logarithm. If one uses a base such as 2 one uses the notation \displaystyle \log_{\,2} for a "base-2 logarithm".
Example 3
- \displaystyle \log_{\,2} 8 = 3\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8.
- \displaystyle \log_{\,2} 2 = 1\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2.
- \displaystyle \log_{\,2} 1024 = 10\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024.
- \displaystyle \log_{\,2}\frac{1}{4} = -2\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2} = \frac{1}{4}.
One deals with logarithms which have other bases in the same way.
Example 4
- \displaystyle \log_{\,3} 9 = 2\quad because \displaystyle 3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9.
- \displaystyle \log_{\,5} 125 = 3\quad because \displaystyle 5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125.
- \displaystyle \log_{\,4} \frac{1}{16} = -2\quad because \displaystyle 4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2} = \frac{1}{16}.
- \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 one uses the notation lg, or simply log, which appears on many calculators.
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 one uses the notation ln instead of \displaystyle \log_{\,e}.
Example 5
- \displaystyle \ln 10 \approx 2{,}3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}3\,}} \approx 10.
- \displaystyle \ln e = 1\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e.
- \displaystyle \ln\frac{1}{e^3} = -3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}} = \frac{1}{e^3}.
- \displaystyle \ln 1 = 0\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
- If \displaystyle y= e^{\,a} then \displaystyle a = \ln y.
- \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5
- \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x
Most advanced calculators usually 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 goes much faster to perform 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
and we then know that \displaystyle 10^{\,3\textrm{.}2765} \approx 1890 (i.e. \displaystyle \lg 1890 \approx 3\textrm{.}2765) thus we have managed to calculate the product
and this just by adding together exponents \displaystyle 1\textrm{.}5441 and \displaystyle 1\textrm{.}7324.
This is an example of a logarithmic law which says that
This stems from the fact that on the one hand,
and on the other hand,
By exploiting the laws of exponents in this way can we obtain the corresponding laws of logarithms:
The laws of logarithms apply regardless of base.
Example 7
- \displaystyle \lg 4 + \lg 7 = \lg(4 \cdot 7) = \lg 28
- \displaystyle \lg 6 - \lg 3 = \lg\frac{6}{3} = \lg 2
- \displaystyle 2 \cdot \lg 5 = \lg 5^2 = \lg 25
- \displaystyle \lg 200 = \lg(2 \cdot 100) = \lg 2 + \lg 100 = \lg 2 + 2
Example 8
- \displaystyle \lg 9 + \lg 1000 - \lg 3 + \lg 0{,}001 = \lg 9 + 3 - \lg 3 - 3 = \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3
- \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(} - \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(} - \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 sometimes can be a good idea to express a logarithm as a logarithm having another base.
Example 9
- Express \displaystyle \lg 5 as a natural logarithm.
By definition, the \displaystyle \lg 5 is a number that satisfies the equality Vorlage:Displayed math Take the natural logarithm ( ln ) of both sides. Vorlage:Displayed math 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 Vorlage:Displayed math Now divide both sides by \displaystyle \ln 10 giving the answer Vorlage:Displayed math - 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 Vorlage:Displayed math and taking the 10-logarithm (lg) of both sides, one gets Vorlage:Displayed math Since \displaystyle \lg a^b = b \lg a one gets \displaystyle \lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2 and the right-hand side can be simplified to \displaystyle \lg 100 = 2. This gives the equality Vorlage:Displayed math Finally, dividing by \displaystyle \lg 2 gives that Vorlage:Displayed math
The general formula for changing from one base \displaystyle a to another base \displaystyle b can be derived in the same way
If one wants to change the base of a power, one can do this by using logarithms. For instance, if we want to write \displaystyle 2^5 using the base 10 one first writes 2 as a power with the base 10; Vorlage:Displayed math
and then using one of the laws of exponents Vorlage:Displayed math
Example 10
- Write \displaystyle 10^x using the base e.
First, we write 10 as a power of e, Vorlage:Displayed math and then use the laws of exponents Vorlage:Displayed math - Write \displaystyle e^{\,a} using the base 10.
The number \displaystyle e can be written as \displaystyle e=10^{\lg e} and therefore Vorlage:Displayed math
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 much time studying logarithms.
Logarithms usually are dealt with summarily 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 studies or need more detailed explanations consider the following references:
Learn more about logarithms in English Wikipedia
Learn more about the number e in The MacTutor History of Mathematics archive
Useful web sites
Experiment with logarithms and powers
Help the frog to jump onto his water-lily leaf in the "log" game
