MATHHX B

MATHHX B

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7.2 Logaritmefunktioner

Nu hvor vi har styr på omvendte funktioner er vi klar til at lære om logaritmer. En logaritmefunktion er en omvendt funktion til en eksponentialfunktion. Altså den omvendte funktion til en funktion på formen \(f(x)=a^x\). På HHX er vi interesserede i to logaritmefunktioner:

  • Definition 7.2.1
    Titalslogaritmen defineres som den omvendte funktion til \(f(x)=10^x\) og betegnes med \(\log (x)\):

    (-tikz- diagram)

    Den naturlige logaritmefunktion defineres som den omvendte funktion til \(f(x)=e^x\) og betegnes med \(\ln (x)\) (det er et ”L” ikke et ”i”):

    (-tikz- diagram)

Skrivemåden ”\(\ln (x)\)” udtales ”L N til x”, ”L N af x”. Tilsvarende for \(\log (x)\).

Øvelse 7.2.1

Sig højt

  • a) \(\ln (x)\)

  • b) \(\log (x)\)

Løsning 7.2.1

  • a) Sagde du ”LN af x”, ”LN til x” eller ”LN x”? Jeg håber ikke du sagde ”L af N”. Det er nemlig en almindelig fejl.

  • b) Sagde du ”log af x”, ”log til x” eller ”log x”?

Øvelse 7.2.2

Vi har indført \(\log (x)\) og \(\ln (x)\) som omvendte funktioner til \(f(x)=10^x\) og \(f(x)=e^x\).

  • a) Hvordan kan vi være sikker på at \(f(x)=10^x\) og \(f(x)=e^x\) overhovedet har omvendte funktioner?

Løsning 7.2.2

  • a) I sidste afsnit så vi (øvelse 7.1.5) at alle eksponentielle funktioner er invertible. Altså må \(f(x)=10^x\) og \(f(x)=e^x\) også være invertible, dvs. de har omvendte funktioner.

  • Eksempel 7.2.1
    Vi vil bestemme \(\log (100)\). Vi husker at \(\log (x)\) er den omvendte funktion til funktionen \(f(x)=10^x\). Vi har altså

    (-tikz- diagram)

    Hvad mon \(x\) kan være? Vi skal have at \(10^x=100\), så \(x\) må være \(2\). Vi konkluderer, at \(\log (100)=2.\)

Øvelse 7.2.3

Bestem ved hoveregning:

  • a) \(\log (1000)\)

  • b) \(\log (10)\)

  • c) \(\ln (e^3)\)

  • d) \(\ln (e)\)

  • e) \(\ln (1)\)

  • f) \(\log (0{,}1)\) (svær)

Løsning 7.2.3

  • a) \(\log (1000)=3\)

  • b) \(\log (10)=1\)

  • c) \(\ln (e^3)=3\)

  • d) \(\ln (e)=1\)

  • e) \(\ln (1)=0\)

  • f) \(\log (0{,}1)=-1\)

  • Eksempel 7.2.2
    Vi vil regne \(e^{\ln (17)}\). Dvs. vi skal finde ud af hvad der sker når vi har \(17\), bruger \(\ln \) og derefter bruger funktionen \(f(x)=e^x\). Da \(f(x)=e^x\) er den omvendte funktion til \(\ln (x)\) må resultatet være \(17\). Altså

    \[e^{\ln (17)}=17\]

Øvelse 7.2.4

Beregn ved hovedregning:

  • a) \(e^{\ln (5)}\)

  • b) \(\ln (e^3)\)

  • c) \(\log (10^7)\)

  • d) \(10^{\log (2)}\)

Løsning 7.2.4

  • a) \(e^{\ln (5)}=5\)

  • b) \(\ln (e^3)=3\)

  • c) \(\log (10^7)=7\)

  • d) \(10^{\log (2)}=2\)

Vi kan løse ligninger med logaritmer, hvis bare vi husker, at de er defineret som omvendte funktioner.

  • Eksempel 7.2.3
    Vi vil løse ligningen:

    \[2\cdot \log (x)-3=2\]

    Vi lægger \(3\) til på begge sider:

    \[2\cdot \log (x)=5\]

    og dividere med \(2\) på begge sider

    \[\log (x)=2{,}5\]

    Vi husker nu \(\log (x)\) er omvendt funktion til \(f(x)=10^x\), så vi bruger \(f(x)=10^x\) på begge sider:

    \[10^{\log (x)}=10^{2{,}5}\]

    På venstresiden bliver det bare \(x\) og højresiden taster vi i GeoGebra:

    \[x=316{,}23\]

Øvelse 7.2.5

Løs ligningerne:

  • a) \(2\log (x)=6\)

  • b) \(\ln (x)=2\)

  • c) \(2+\ln (x)=2\cdot \ln (x)+1\)

Løsning 7.2.5

  • a) \(x=1000\)

  • b) \(x=7{,}39\)

  • c) \(x=e=2{,}72\)

Øvelse 7.2.6

Lad \(f(x)=-1\) og \(g(x)=\ln (x)+1\)

  • a) Beregn skæringspunktet mellem \(f\) og \(g\).

Løsning 7.2.6

  • a) \((0{,}14;-1)\)

Grafer for logaritmefunktioner

Graferne for \(\log (x)\) og \(\ln (x)\) ser således ud:

(-tikz- diagram) (-tikz- diagram)

Øvelse 7.2.7

Graferne for \(f(x)=10^x\) og \(f(x)=e^x\) ser således ud:

(-tikz- diagram) (-tikz- diagram)

  • a) Brug grafen for \(f(x)=10^x\) til at argumentere for grafen for \(\log (x)\).

  • b) Brug grafen for \(f(x)=e^x\) til at argumentere for grafen for \(\ln (x)\).

Løsning 7.2.7

  • a) Vi husker at \(\log (x)\) er omvendt funktion til \(f(x)=10^x\), så dens graf fremkommer ved at spejle i \(y=x\).

    (-tikz- diagram)

  • b) Vi husker at \(\ln (x)\) er omvendt funktion til \(f(x)=e^x\), så dens graf fremkommer ved at spejle i \(y=x\).

    (-tikz- diagram)

Kigger man på graferne for \(\log (x)\) og \(\ln (x)\) ser det ud som at definitionsmængden for begge logaritmefunktioner er \(]0;\infty [\) og at værdimængden er \(]-\infty ;\infty [\). Dette er korrekt.

Øvelse 7.2.8

Vi husker at definitionsmængden for eksponentielle funktioner er \(]-\infty ;\infty [\) og værdimængden er \(]0;\infty [\).

  • a) Brug definitions og værdimængden for eksponentielle funktioner til at argumentere for definitions og værdimængden for logaritmefunktioner.

Løsning 7.2.8

  • a) Da logaritmefunktioner er omvendte funktioner til eksponentielle funktioner vil definitions og værdimængden svarer til definitions og værdimængden for eksponentielle funktioner, bare byttet rundt. Altså bliver definitionsmængden for logaritmefunktionerne \(]0;\infty [\) og værdimængden bliver \(]-\infty ;\infty [\).

Øvelse 7.2.9 (Svær)

Som det fremgår af graferne har både \(\log (x)\) og \(\ln (x)\) nulpunkt i \(x=1\).

  • a) Gør rede for at \(\log (x)\) har nulpunkt i \(x=1\)

  • b) Gør rede for at \(\ln (x)\) har nulpunkt i \(x=1\)

Løsning 7.2.9

  • a) Nulpunkterne bestemmes ved at løse ligningen \(\log (x)=0\). Vi bruger funktion \(f(x)=10^x\) på begge sider:

    \[10^{\log (x)}=10^0\]

    Da \(\log (x)\) er omvendt funktion til \(f(x)=10^x\) fås

    \[x=10^0\]

    dvs.

    \[x=1\]

  • b) Nulpunkterne bestemmes ved at løse ligningen \(\ln (x)=0\). Vi bruger funktion \(f(x)=e^x\) på begge sider:

    \[e^{\log (x)}=e^0\]

    Da \(\ln (x)\) er omvendt funktion til \(f(x)=1e^x\) fås

    \[x=e^0\]

    dvs.

    \[x=1\]

Ekstra

Der gælder følgende regneregler for logaritmer:

\(\begin {array}{| l | l |} \hline \text {Den naturlige logaritme} & \text {Titalslogaritmen}\\ \hline \ln (1)=0 & \log (1)=0\\ \ln (e)=1 & \log (10)=1\\ \ln (a\cdot b) = \ln (a) + \ln (b) & \log (a\cdot b) = \log (a) + \log (b)\\ \ln (\frac {a}{b})=\ln (a)-\ln (b) & \log (\frac {a}{b})=\log (a)-\log (b)\\ \ln (a^p)=p\cdot \ln (a) & \log a^p = p \cdot \log (a)\\ \hline \end {array}\)

Øvelse 7.2.10

Regn med hovedregning og regneregler.

  • a) \(\log (1)\)

  • b) \(\log (10000)\)

  • c) \(\ln (e^{27})\)

  • d) \(\ln (\frac {1}{e})\)

  • e) \(\log (200)-\log (20)\)

  • f) \(\ln (\frac {1}{2}e)+\ln (2e)\)

Løsning 7.2.10

  • a) \(\log (1)=0\)

  • b) \(\log (10000)=4\)

  • c) \(\ln (e^{27})=27\)

  • d) \(\ln (\frac {1}{e})=-1\)

  • e) \(\log (200)-\log (20)=1\)

  • f) \(\ln (\frac {1}{2}e)+\ln (2e)=2\)

Ud over de viste regneregler gælder der også:

\[\ln (x)=\frac {\log (x)}{\log (e)}\quad \text {og}\quad \log (x)=\frac {\ln (x)}{\ln (10)}\]

Vi kan altså finde \(\ln (x)\) ved dividere \(\log (x)\) med konstanten \(\log (e)\). Dvs. \(\ln (x)\) er altså den samme funktion som \(\log (x)\) bare \(\ln (10)\) gange mindre.

Øvelse 7.2.11

Regn i GeoGebra, men uden at bruge titalslogaritmen.

  • a) \(\log (7)\)

Løsning 7.2.11

  • a) \(\log (7)=0{,}85\)