3.3 Logarithmen

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{{Vald flik|[[3.3 Logaritmer|Teori]]}}
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{{Vald flik|[[3.3 Logaritmer|Theory]]}}
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{{Ej vald flik|[[3.3 Övningar|Övningar]]}}
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{{Ej vald flik|[[3.3 Övningar|Exercise]]}}
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'''Innehåll:'''
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'''Content:'''
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*Logaritmer
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* Logarithms
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*Logaritmlagar
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* Fundamental Laws of Logarithms
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{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned :
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*Känna till begreppen bas och exponent.
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*The concepts of base and exponent.
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*Känna till beteckningarna <math>\ln</math>, <math>\lg</math>, <math>\log</math> och <math>\log_{a}</math>.
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*The meaning of the notation <math>\ln</math>, <math>\lg</math>, <math>\log</math> and <math>\log_{a}</math>.
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*Beräkna enkla logaritmuttryck med hjälp av logaritmens definition.
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*To calculate simple logarithmic expressions using the definition of a logarithm.
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*Logaritmen är bara definierad för positiva tal.
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*That logarithms are only defined for positive numbers.
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*Känna till talet <math>e</math>.
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* the meaning of the number <math>e</math>.
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*Hantera logaritmlagarna i förenkling av logaritmuttryck.
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* To use the laws of logarithms to simplify logarithmic expressions.
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*Veta när logaritmlagarna är giltiga.
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* To know when the laws of logarithms are valid.
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*Uttrycka en logaritm i termer av en logaritm med en annan bas.
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* To express a logarithm in terms of a logarithm with a different base.
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*Lösa ekvationer som innehåller exponentialuttryck och som med logaritmering leder till förstagradsekvationer.
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* To solve equations that contain powers and by taking logarithms obtain an equation of the first degree.
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*Avgöra vilket av två logaritmuttryck som är störst baserat på jämförelse av bas/argument.
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* To determining which of two logarithmic expressions is the largest by means of a comparison of bases / arguments.
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== Logaritmer med basen 10 ==
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==Logarithms to the base 10 ==
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Man använder gärna potenser med basen <math>10</math> för att skriva stora och små tal, t.ex.
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We often use powers with base <math>10</math> to represent large and small numbers, for example,
{{Fristående formel||<math>\begin{align*}
{{Fristående formel||<math>\begin{align*}
Zeile 38: Zeile 38:
\end{align*}</math>}}
\end{align*}</math>}}
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Om man enbart betraktar exponenten skulle man i stället kunna säga att
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If one only considers the exponents one can state that
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:::"exponenten för 1000 är 3", eller
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:::"exponent for 1000 is 3", or
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:::"exponenten för 0,01 är -2".
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:::" exponent for 0,01 is -2".
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Precis så är ''logaritmer'' definierade. Man uttrycker sig på följande sätt:
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This how logarithms are defined. One formalises this as follows:
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:::"''logaritmen'' för 1000 är 3", vilket skrivs <math>\lg 1000 = 3</math>,
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:::"''The logarithm'' of 1000 is 3", which is written as <math>\lg 1000 = 3</math>,
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:::"''logaritmen'' för 0,01 är -2", vilket skrivs <math>\lg 0{,}01 = -2</math>.
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:::"''The logarithm'' of 0,01 är -2", which is written as <math>\lg 0{,}01 = -2</math>.
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Mer allmänt kan man uttrycka sig:
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More generally, one says:
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:::Logaritmen av ett tal <math>y</math> betecknas med <math>\lg y</math> och är den exponent som ska stå i den blåa rutan i likheten
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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
{{Fristående formel||<math>10^{\ \bbox[#AAEEFF,2pt]{\,\phantom{a}\,}} = y\,\mbox{.} </math>}}
{{Fristående formel||<math>10^{\ \bbox[#AAEEFF,2pt]{\,\phantom{a}\,}} = y\,\mbox{.} </math>}}
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Notera här att <math>y</math> måste vara ett positivt tal för att logaritmen <math>\lg y</math> ska vara definerad, eftersom det inte finns någon potens av 10 som blir negativ eller noll.
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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 .
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
<ol type="a">
<ol type="a">
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<li><math>\lg 100000 = 5\quad</math> eftersom <math>
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<li><math>\lg 100000 = 5\quad</math> because <math>
10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}}
10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}}
= 100\,000</math>.</li>
= 100\,000</math>.</li>
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<li><math>\lg 0{,}0001 = -4\quad</math> eftersom <math>
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<li><math>\lg 0{,}0001 = -4\quad</math> because <math>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}}
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}}
= 0{,}0001</math>.</li>
= 0{,}0001</math>.</li>
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<li><math>\lg \sqrt{10} = \frac{1}{2}\quad</math> eftersom <math>
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<li><math>\lg \sqrt{10} = \frac{1}{2}\quad</math> because <math>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}</math>.</li>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}</math>.</li>
<li><math>\lg 1 = 0\quad</math> eftersom <math>
<li><math>\lg 1 = 0\quad</math> eftersom <math>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li>
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<li><math>\lg 10^{78} = 78\quad</math> eftersom <math>
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<li><math>\lg 10^{78} = 78\quad</math> because <math>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}}
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}}
= 10^{78}</math>.</li>
= 10^{78}</math>.</li>
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<li><math>\lg 50 \approx 1{,}699\quad</math> eftersom <math>
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<li><math>\lg 50 \approx 1{,}699\quad</math> because <math>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1{,}699\,}} \approx 50</math>.</li>
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1{,}699\,}} \approx 50</math>.</li>
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<li><math>\lg (-10)</math> existerar inte eftersom <math>
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<li><math>\lg (-10)</math> does not exist because <math>
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10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}}</math> aldrig kan bli -10 oavsett hur <math>a</math> väljs.</li>
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10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}}</math> can never be -10 regardless of how <math>a</math> is chosen.</li>
</ol>
</ol>
</div>
</div>
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I det näst sista exemplet kan man snabbt inse att <math>\lg 50</math> måste ligga någonstans mellan 1 och 2 eftersom <math>10^1 < 50 < 10^2</math>, men för att få fram ett mer exakt värde på det irrationella talet <math>\lg 50 = 1{,}69897\ldots</math> behövs i praktiken en miniräknare (eller tabell.)
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In the penultimate example, one can easily understand that <math>\lg 50</math> must lie somewhere between 1 and 2 since <math>10^1 < 50 < 10^2</math>, but to obtain a more precise value of the irrational number <math>\lg 50 = 1{,}69897\ldots</math> one needs in practice, a calculator (or table.)
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
<ol type="a">
<ol type="a">
Zeile 93: Zeile 93:
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== Olika baser ==
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== Different bases ==
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Man kan tänka sig logaritmer som använder en annan bas än 10 (utom 1!). Man måste då tydligt ange vilket tal man använder som bas för logaritmen. Använder man t.ex. 2 som bas skriver man <math>\log_{\,2}</math> för "2-logaritmen".
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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 <math>\log_{\,2}</math> for "2-logaritmen".
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
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<li><math>\log_{\,2} 8 = 3\quad</math> eftersom <math>
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<li><math>\log_{\,2} 8 = 3\quad</math> because <math>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8</math>.</li>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8</math>.</li>
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<li><math>\log_{\,2} 2 = 1\quad</math> eftersom <math>
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<li><math>\log_{\,2} 2 = 1\quad</math> because <math>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2</math>.</li>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2</math>.</li>
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<li><math>\log_{\,2} 1024 = 10\quad</math> eftersom <math>
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<li><math>\log_{\,2} 1024 = 10\quad</math> because <math>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024</math>.</li>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024</math>.</li>
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<li><math>\log_{\,2}\frac{1}{4} = -2\quad</math> eftersom <math>
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<li><math>\log_{\,2}\frac{1}{4} = -2\quad</math> because <math>
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2}
2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2}
= \frac{1}{4}</math>.</li>
= \frac{1}{4}</math>.</li>
Zeile 113: Zeile 113:
</div>
</div>
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På samma sätt fungerar logaritmer i andra baser.
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One deals with logarithms which have other bases in the same way.
 +
 
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
-
<li><math> \log_{\,3} 9 = 2\quad</math> eftersom <math>
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<li><math> \log_{\,3} 9 = 2\quad</math> because <math>
3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9</math>.</li>
3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9</math>.</li>
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<li><math> \log_{\,5} 125 = 3\quad</math> eftersom <math>
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<li><math> \log_{\,5} 125 = 3\quad</math> because <math>
5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125</math>.</li>
5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125</math>.</li>
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<li><math> \log_{\,4} \frac{1}{16} = -2\quad</math> eftersom <math>
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<li><math> \log_{\,4} \frac{1}{16} = -2\quad</math> because <math>
4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2}
4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2}
= \frac{1}{16}</math>.</li>
= \frac{1}{16}</math>.</li>
<li><math> \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad
<li><math> \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad
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</math> eftersom <math>b^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-1/2\,}}
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</math> as the <math>b^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-1/2\,}}
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= \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}}</math> (om <math>b>0</math> och <math>b\not=1</math>).</li>
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= \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}}</math> (if <math>b>0</math> and <math>b\not=1</math>).</li>
</ol>
</ol>
</div>
</div>
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Om basen 10 används, skriver man sällan <math>\log_{\,10}</math>, utan som vi tidigare sett lg, eller enbart log, vilket förekommer på många miniräknare.
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If the base 10 is, used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen one uses the notation lg, or simply log, which appears on many calculators.
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== Naturliga logaritmer ==
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==The natural logarithms ==
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I praktiken är det två baser som oftast används för logaritmer, förutom 10 även talet <math>e</math> <math>({}\approx 2{,}71828 \ldots\,)</math>. Logaritmer med basen ''e'' kallas ''naturliga logaritmer'' och skrivs ln i stället för <math>\log_{\,e}</math>.
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In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2{,}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and one uses the notation ln instead of <math>\log_{\,e}</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
<ol type="a">
<ol type="a">
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<li><math> \ln 10 \approx 2{,}3\quad</math> eftersom <math>
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<li><math> \ln 10 \approx 2{,}3\quad</math> because <math>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}3\,}} \approx 10</math>.</li>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}3\,}} \approx 10</math>.</li>
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<li><math> \ln e = 1\quad</math> eftersom <math>
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<li><math> \ln e = 1\quad</math> because <math>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li>
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<li><math> \ln\frac{1}{e^3} = -3\quad</math> eftersom <math>
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<li><math> \ln\frac{1}{e^3} = -3\quad</math> because <math>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}}
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}}
= \frac{1}{e^3}</math>.</li>
= \frac{1}{e^3}</math>.</li>
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<li><math> \ln 1 = 0\quad</math> eftersom <math>
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<li><math> \ln 1 = 0\quad</math> because <math>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li>
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li>
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<li>Om <math>y= e^{\,a}</math> så är <math>a = \ln y</math>.</li>
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<li>Om <math>y= e^{\,a}</math> then <math>a = \ln y</math>.</li>
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5</math></li>
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5</math></li>
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x</math></li>
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x</math></li>
Zeile 158: Zeile 159:
</div>
</div>
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På de flesta mer avancerade miniräknare finns vanligtvis knappar för 10-logaritmer och naturliga logaritmer.
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Most advanced calculators usually have buttons for 10-logarithms and natural logarithms.
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== Logaritmlagar ==
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== Laws of Logarithms ==
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Mellan år 1617 och 1624 publicerade Henry Biggs en logaritmtabell av alla heltal upp till 20 000 och år 1628 utökade Adriaan Vlacq tabellen till alla heltal upp till 100 000. Anledningen till att man la ned så enormt mycket arbete på sådana tabeller är att man med hjälp av logaritmer kan multiplicera ihop tal bara genom att addera ihop deras logaritmer (addition går mycket snabbare att utföra än multiplikation).
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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).
<div class="exempel">
<div class="exempel">
-
'''Exempel 6'''
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''' Example 6'''
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Beräkna <math>\,35\cdot 54</math>.
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Calculate <math>\,35\cdot 54</math>.
<br>
<br>
<br>
<br>
-
Om vi vet att <math>35 \approx 10^{\,1{,}5441}</math> och <math>54 \approx 10^{\,1{,}7324}</math> (dvs. <math>\lg 35 \approx 1{,}5441</math> och <math>\lg 54 \approx 1{,}7324</math>) då kan vi räkna ut att
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If we know that <math>35 \approx 10^{\,1{,}5441}</math> and <math>54 \approx 10^{\,1{,}7324}</math> (i.e. <math>\lg 35 \approx 1{,}5441</math> and <math>\lg 54 \approx 1{,}7324</math>) then we can calculate that
{{Fristående formel||<math>
{{Fristående formel||<math>
Zeile 178: Zeile 179:
= 10^{\,3{,}2765}</math>}}
= 10^{\,3{,}2765}</math>}}
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och vet vi sedan att <math>10^{\,3{,}2765} \approx 1890</math> (dvs. <math>\lg 1890 \approx 3{,}2765</math>) så har vi lyckats beräkna produkten
+
and we then know that <math>10^{\,3{,}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3{,}2765</math>) thus we have managed to calculate the product
{{Fristående formel||<math>35 \cdot 54 = 1890</math>}}
{{Fristående formel||<math>35 \cdot 54 = 1890</math>}}
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och detta bara genom att addera ihop exponenterna <math>1{,}5441</math> och <math>1{,}7324</math>.
+
and this just by adding together exponents <math>1{,}5441</math> and <math>1{,}7324</math>.
</div>
</div>
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Detta är ett exempel på en logaritmlag som säger att
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This is an example of a logarithmic law which say that
{{Fristående formel||<math>\log (ab) = \log a + \log b</math>}}
{{Fristående formel||<math>\log (ab) = \log a + \log b</math>}}
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och som följer av att å ena sidan är
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This stems from the fact that on the one hand,
{{Fristående formel||<math>
{{Fristående formel||<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{potenslagarna} \right\}
+
= \left\{ \mbox{laws of powers} \right\}
= 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}}
= 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}}
-
och å andra sidan är
+
and on the other hand,
{{Fristående formel||<math>
{{Fristående formel||<math>
a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}}
a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}}
-
Genom att utnyttja potenslagarna på detta sätt kan vi få fram motsvarande ''logaritmlagar'':
+
By exploiting the laws of powers in this way can we obtain the corresponding ''laws of logarithms'':
<div class="regel">
<div class="regel">
Zeile 211: Zeile 212:
</div>
</div>
-
Logaritmlagarna gäller oavsett bas.
+
The laws of logarithms apply regardless of base.
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
+
''' Example 7'''
<ol type="a">
<ol type="a">
Zeile 259: Zeile 260:
-
== Byte av bas ==
+
== Changing the base ==
-
Ibland kan det vara bra att kunna uttrycka en logaritm som en logaritm av en annan bas.
+
It sometimes can be a good idea to express a logarithm as a logarithm having another base.
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
-
<li>Uttryck <math>\lg 5</math> i naturliga logaritmen.
+
<li> Express <math>\lg 5</math> as a natural logarithm.
<br>
<br>
<br>
<br>
-
Per definition är <math>\lg 5</math> det tal som uppfyller likheten
+
By definition, the <math>\lg 5</math> is a number that satisfies the equality
{{Fristående formel||<math>10^{\lg 5} = 5\,\mbox{.}</math>}}
{{Fristående formel||<math>10^{\lg 5} = 5\,\mbox{.}</math>}}
-
Logaritmera båda led med ln (naturliga logaritmen)
+
Take the natural logarithm ( ln ) of both sides.
{{Fristående formel||<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}
{{Fristående formel||<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}}
-
Med hjälp av logaritmlagen <math>\ln a^b = b \ln a</math> kan vänsterledet skrivas som <math>\lg 5 \cdot \ln 10</math> och likheten blir
+
With the help of the logarithm law <math>\ln a^b = b \ln a</math> the left-hand side can written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes
{{Fristående formel||<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}
{{Fristående formel||<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}}
-
Dela nu båda led med <math>\ln 10</math> så får vi svaret
+
  Now divide both sides by <math>\ln 10</math> giving the answer
{{Fristående formel||<math>
{{Fristående formel||<math>
\lg 5 = \frac{\ln 5}{\ln 10}
\lg 5 = \frac{\ln 5}{\ln 10}
Zeile 286: Zeile 287:
</math>}}</li>
</math>}}</li>
-
<li> Uttryck 2-logaritmen för 100 i 10-logaritmen lg.
+
<li> Express the 2-logarithm of 100 as a 10-logarithm lg.
<br>
<br>
<br>
<br>
-
Om vi skriver upp sambandet som definierar <math>\log_2 100</math>
+
Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies
-
{{Fristående formel||<math>2^{\log_{\scriptstyle 2} 100} = 100</math>}}
+
{{Fristående formel||<math>2^{\log_{\scriptstyle 2} 100} = 100</math>}}
-
och logaritmerar båda led med 10-logaritmen (lg) så får vi att
+
and take the 10-logarithm (lg) of both sides, one get s
{{Fristående formel||<math>
{{Fristående formel||<math>
\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}}
\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}}
-
Eftersom <math>\lg a^b = b \lg a</math> så är <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math> och högerledet kan förenklas till <math>\lg 100 = 2</math>. Detta ger oss likheten
+
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
{{Fristående formel||<math>
{{Fristående formel||<math>
\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}}
\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}}
-
Division med <math>\lg 2</math> ger slutligen att
+
Finally dividing by <math>\lg 2</math> gives that
{{Fristående formel||<math>
{{Fristående formel||<math>
\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2}
\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2}
Zeile 308: Zeile 309:
</div>
</div>
-
Den allmänna formeln för byte från en bas <math>a</math> till en bas <math>b</math> kan härledas på samma sätt
+
The general formula for changing from one base <math>a</math> to another base <math>b</math> can be derived in the same way
{{Fristående formel||<math>
{{Fristående formel||<math>
Zeile 315: Zeile 316:
\,\mbox{.}</math>}}
\,\mbox{.}</math>}}
-
Vill man byta bas i en potens kan man göra detta med hjälp av logaritmer. Om man exempelvis vill skriva <math> 2^5 </math> med basen 10 så skriver man först om 2 med basen 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;
{{Fristående formel||<math>2 = 10^{\lg 2}</math>}}
{{Fristående formel||<math>2 = 10^{\lg 2}</math>}}
-
och utnyttjar sedan en av potenslagarna
+
and then using one of the laws of powers
{{Fristående formel||<math>
{{Fristående formel||<math>
2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2}
2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2}
Zeile 324: Zeile 325:
<div class="exempel">
<div class="exempel">
-
'''Exempel 10'''
+
''' Example 10'''
<ol type="a">
<ol type="a">
-
<li>Skriv <math> 10^x </math> med basen ''e''.
+
<li>Write <math> 10^x </math> using the base ''e''.
<br>
<br>
<br>
<br>
-
Först skriver vi 10 som en potens av ''e'',
+
   First, we write 10 as a power of ''e'',
{{Fristående formel||<math>10 = e^{\ln 10}</math>}}
{{Fristående formel||<math>10 = e^{\ln 10}</math>}}
-
och använder sedan potenslagarna
+
and then use the laws of powers
{{Fristående formel||<math>
{{Fristående formel||<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{,}3 x}\,\mbox{.}</math>}}</li>
\approx e^{2{,}3 x}\,\mbox{.}</math>}}</li>
-
<li>Skriv <math>e^{\,a}</math> med basen 10.
+
<li> Write <math>e^{\,a}</math> using the base 10.
<br>
<br>
<br>
<br>
-
Talet <math>e</math> kan vi skriva som <math>e=10^{\lg e}</math> och därför är
+
The number <math>e</math> can be writen as <math>e=10^{\lg e}</math> and therefore
{{Fristående formel||<math>
{{Fristående formel||<math>
e^a = (10^{\lg e})^a
e^a = (10^{\lg e})^a
Zeile 350: Zeile 351:
-
[[3.3 Övningar|Övningar]]
+
[[3.3 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
-
'''Råd för inläsning'''
+
'''Study advice'''
-
'''Grund- och slutprov'''
+
'''The basic and final tests'''
-
Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
+
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.
-
'''Tänk på att:'''
+
'''Keep in mind that: '''
-
Du kan behöva lägga ner mycket tid på logaritmer.
+
You may need to spend much time studying logarithms.
-
Logaritmer brukar behandlas översiktligt i gymnasiet. Därför brukar många högskolestudenter stöta på problem när det gäller att räkna med logaritmer.
+
Logarithms usually are dealt with summarily in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.
-
'''Lästips'''
+
'''Reviews'''
-
För dig som vill fördjupa dig ytterligare eller behöver en längre förklaring
+
For those of you who want to deepen your studies or need more detailed explanations consider the following references:
-
[http://en.wikipedia.org/wiki/Logarithm Läs mer om logaritmer på engelska Wikipedia]
+
[http://en.wikipedia.org/wiki/Logarithm Learn more about logarithms in English Wikipedia ]
-
[http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Läs mer om Talet ''e'' i The MacTutor History of Mathematics archive]
+
[http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Learn more about the number ''e'' in The MacTutor History of Mathematics archive ]
'''Länktips'''
'''Länktips'''
-
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogGraph/logGraph.html Experimentera med logaritmer och potenser]
+
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogGraph/logGraph.html Experiment with logarithms and powers ]
-
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogConcentration/LogConcentration.htm Spela logaritm Memory]
+
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogConcentration/LogConcentration.htm Play logarithm Memory ]
-
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/logger.htm Hjälp grodan hoppa till sitt näckrosblad i "log"-spelet ]
+
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/logger.htm Help the frog to jump onto his water-lily leaf in the "log" game]
</div>
</div>

Version vom 10:25, 14. Jul. 2008

 

Vorlage:Vald flik Vorlage:Ej vald flik

 

Content:

  • 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.
  • To solve equations that contain powers and by taking logarithms obtain an equation of the first degree.
  • To determining which of two logarithmic expressions is the largest by means of a comparison of bases / arguments.

Logarithms to the base 10

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

Vorlage:Fristående formel

If one only considers the exponents one can state that

"exponent for 1000 is 3", or
" exponent for 0,01 is -2".

This 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 är -2", which is written as \displaystyle \lg 0{,}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

Vorlage:Fristående formel

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{,}0001 = -4\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} = 0{,}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 eftersom \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{,}699\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1{,}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, but to obtain a more precise value of the irrational number \displaystyle \lg 50 = 1{,}69897\ldots one needs in practice, 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

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 "2-logaritmen".

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}.

One deals 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 the \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{,}71828 \ldots\,). Logarithms using the base e are called natural logarithms and one uses the notation ln instead of \displaystyle \log_{\,e}.

Example 5

  1. \displaystyle \ln 10 \approx 2{,}3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}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. Om \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 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{,}5441} and \displaystyle 54 \approx 10^{\,1{,}7324} (i.e. \displaystyle \lg 35 \approx 1{,}5441 and \displaystyle \lg 54 \approx 1{,}7324) then we can calculate that

Vorlage:Fristående formel

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

Vorlage:Fristående formel

and this just by adding together exponents \displaystyle 1{,}5441 and \displaystyle 1{,}7324.

This is an example of a logarithmic law which say that

Vorlage:Fristående formel

This stems from the fact that on the one hand,

Vorlage:Fristående formel

and on the other hand,

Vorlage:Fristående formel

By exploiting the laws of powers in this way can we obtain the corresponding laws of logarithms:

Vorlage:Fristående formel

The laws of logarithms apply regardless of base.

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

Exempel 8

  1. \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
  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 sometimes can be a good idea to express a logarithm as a logarithm having another base.

Example 9

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

    By definition, the \displaystyle \lg 5 is a number that satisfies the equality Vorlage:Fristående formel Take the natural logarithm ( ln ) of both sides. Vorlage:Fristående formel With the help of the logarithm law \displaystyle \ln a^b = b \ln a the left-hand side can written as \displaystyle \lg 5 \cdot \ln 10 and the equality becomes Vorlage:Fristående formel   Now divide both sides by \displaystyle \ln 10 giving the answer Vorlage:Fristående formel
  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 Vorlage:Fristående formel and take the 10-logarithm (lg) of both sides, one get s Vorlage:Fristående formel 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:Fristående formel Finally dividing by \displaystyle \lg 2 gives that Vorlage:Fristående formel

The general formula for changing from one base \displaystyle a to another base \displaystyle b can be derived in the same way

Vorlage:Fristående formel

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:Fristående formel

and then using one of the laws of powers Vorlage:Fristående formel

Example 10

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

       First, we write 10 as a power of e, Vorlage:Fristående formel and then use the laws of powers Vorlage:Fristående formel
  2. Write \displaystyle e^{\,a} using the base 10.

    The number \displaystyle e can be writen as \displaystyle e=10^{\lg e} and therefore Vorlage:Fristående formel


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 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


Länktips

Experiment with logarithms and powers

Play logarithm Memory

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

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/3.3_Logarithmen“