MATHHX B
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7.4 Beviser til logaritmer
Formel for fordoblingskonstant
Vi mødte fordoblingskonstanten i afsnittet om eksponentielle funktioner. Men beviset for formlen kræver at man kan løse eksponentielle ligninger, så derfor er beviser vi formlen her.
Vi vil lave to beviser. Det første er simpelt, men ikke så godt som det andet. Det andet ses i Ekstra-afsnittet. Spring dette bevis over, hvis du har mod på det svære.
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Bevis
Grafen for en eksponentiel funktion skærer \(y\)-aksen i \(b\):
Fordoblingskonstanten \(T_2\) er længen af det stykke, vi skal gå ud af \(x\)-aksen for at fordoble funktionsværdien. Hvis vi starter i \(x=0\) og går ud til \(T_2\) på \(x\)-aksen må funktionsværdien i \(x=T_2\) altså være det
dobbelte af \(b\), nemlig \(2b\):
Ifølge tegningen har vi altså:
\[f(T_2)=2b\]
Vi indsætter forskriften:
\[ba^{T_2}=2b.\]
Vi deler med \(b\) på begge sider:
\[a^{T_2}=2.\]
Vi tager den naturlige logaritme på begge sider:
\[\ln (a^{T_2})=\ln (2).\]
Vi bruger reglen \(\ln (a^p)=p\cdot \ln (a)\):
\[T_2\ln (a)=\ln (2).\]
Vi dividerer med \(\ln (a)\) på begge sider:
\[T_2=\frac {\ln (2)}{\ln (a)},\]
og vi har hermed bevist sætning.
Øvelse 7.4.1
Der gælder en tilsvarende sætning for halveringskonstanten.
Løsning 7.4.1
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a) Vis mig det
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b) Vis mig det
Logaritmeregneregler
Vi har ikke skrevet logaritmeregnereglerne om som en sætning, men når vi laver beviser, vil vi altid formulere det vi beviser som en sætning. Så her er en sætning omhandlende en logaritmeregneregel.
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Bevis
Da \(\ln (x)\) er den omvendte funktion til \(e^x\) gælder at:
\[a=e^{\ln (a)}.\]
Vi opløfter i \(p\) på begge sider:
\[a^p=\left (e^{\ln (a)}\right )^p.\]
Vi tager nu den naturlige logaritme på begge sider:
\[\ln (a^p)=\ln \left (\left (e^{\ln (a)}\right )^p\right ).\]
Vi benytter reglen \({(a^p)}^q=a^{p\cdot q}\) på udtrykket \(\left (e^{\ln (a)}\right )^p\):
\[\ln (a^p)=\ln (e^{\ln (a)\cdot p}).\]
Igen benytter vi at \(\ln (x)\) er den omvendte funktion til \(e^x\) og får højresiden til at være:
\[\ln (a^p)=\ln (a)\cdot p,\]
og det er jo det samme som:
\[\ln (a^p)=p\cdot \ln (a).\]
Ekstra
Den skarpe læser har måske iagttaget et problem i forhold til den måde vi har indført fordoblingskonstanten. Vi har defineret den som det stykke, vi skal gå ud for at fordoble funktionen. Men for næsten alle funktionstyper, vil det
stykke afhænge af, hvor vi starter henne. Dvs. der findes ikke noget fast stykke, man kan gå ud for at fordoble funktionsværdien. Det er en særlig egenskab ved eksponentielle funktioner, at de overhovedet har en fordoblingskonstant,
og vi har derfor også brug for at bevise, at de har denne egenskab. Vi vil nu formulere og bevise sætningen med formlen for fordoblingskonstanten på en måde, som ikke tager udgangspunkt i, at fordoblingskonstanten findes:
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Bevis
Lad \(f(x)=ba^x\) være en eksponentiel funktion. Vi vil gerne vise at funktionen har en fordoblingskonstant \(T_2\). En fordoblingskonstant er et tal \(T_2\), som opfylder at
\[f(x+T_2)=2f(x).,\]
for alle \(x\in \Dm (f)\). Læg mærke til at ligningen udtrykker at funktionen bliver dobbelt så stor, når \(x\) vokser med \(T_2\). Vi vælger først en vilkårlig \(x\)-værdi som vi kalder \(x_0\).
Vi går nu op ad \(y\)-aksen indtil vi rammer det dobbelte af \(f(x_0)\), dvs. \(2f(x_0)\):
Vi finder nu den \(x\)-værdi som hører til \(2f(x_0)\). Stykket mellem \(x_0\) og denne ny x-værdi kalder vi for \(T_2\), og den nye \(x\)-værdi er dermed \(x_0+T_2\):
Vi har altså:
\[f(x_0+T_2)=2f(x_0).\]
Vi indsætter forskriften:
\[ba^{x_0+T_2}=2ba^{x_0}.\]
Vi deler med \(b\) på begge sider:
\[a^{x_0+T_2}=2a^{x_0}.\]
Vi deler med \(a^{x_0}\) på begge sider:
\[\frac {a^{x_0+T_2}}{a^{x_0}}=\frac {2a^{x_0}}{a^{x_0}}.\]
Vi bruger potensregnereglen \(\frac {a^p}{a^q}=a^{p-q}\) på venstre side og reducerer på højre:
\[a^{T_2}=2.\]
Vi tager den naturlige logaritme på begge sider:
\[\ln (a^{T_2})=\ln (2).\]
Vi bruger reglen \(\ln (a^x)=x\ln (a)\):
\[T_2\ln (a)=\ln (2).\]
Vi dividerer med \(\ln (a)\) på begge sider:
\[T_2=\frac {\ln (2)}{\ln (a)}.\]
Vi lægger mærke til at \(T_2\) ikke afhænger af, hvilket \(x_0\) vi startede med at vælge. Vi har derfor fundet en konstant \(T_2\) som opfylder at \(f(x+T_2)=2f(x)\) for alle \(x\) i definitionsmængden. Altså har vi vist at at
funktionen har en fordoblingskonstant som er givet ved:
\[T_2=\frac {\ln (2)}{\ln (a)}.\]
Øvelse 7.4.2
Der gælder en tilsvarende sætning for halveringskonstanten.
Løsning 7.4.2
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a) Vis mig det
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b) Vis mig det