Leerdoelen
•Je leert hoe je logaritmische vergelijkingen moet oplossen met behulp van logaritmische rekenregels.
Opdracht 1: Oplossen van een^3\log^3^3^3^3^3
Opdracht: Los exact op:^3\log(5x)+^3\log(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+^3^{}(6)=4^3\log(5x)+^3^{\prime}(6)=4^3\log(5x)+^3(6)=4^3\log(5x)+(6)=4^3\log(5x)(6)=4^3\log(5x(6)=4^3\log(5x)(6)=4^3\log(5x)+(6)=4^3\log(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4(5x)+\log_3(6)=4^{}(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4^3^{}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3^{\prime\log}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3^{\prime}(5x)+\log_3(6)=4^3(5x)+\log_3(6)=4(5x)+\log_3(6)=4 \log_3(5x) + \log_3(6) = 4 (5x)+\log_3(6)=4
Omdat beide logaritmes dezelfdezijn, kunnen we ze samenvoegen. Dit doen we door de argumenten te vermenigvuldigen: ^3\log(5x\cdot6)=^3\log(30x)=4^3\log(5x\cdot6)=^3\log(30x)=^3\log(5x\cdot6)=^3\log(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=^3(30x)^3\log(5x\cdot6)=(30x)^3\log(5x\cdot6)=\log_3(30x)^3(5x\cdot6)=\log_3(30x)^3(5x\cdot6)=\log_3(30x)^3(5x\cdot6)=\log_3(30x)^3(5x\cdot6)=\log_3(30x)^3(5x\cdot6)=\log_3(30x)(5x\cdot6)=\log_3(30x)
Nu kunnen we de logaritme weghalen door elke zijde als exponent van de basis (hierte nemen: We weten 3^4=81,(3^4=81,dus:
Deel beide zijden door x=\frac{81}{30}\text{ of }x=2\frac{7}{10}x=\frac{81}{30}\text{ of }x=\frac{7}{10}x=\frac{81}{30}\text{ of }x=\frac71x=\frac{81}{30}\text{ of }x=\frac{7}{\placeholder{}}x=\frac{81}{30}\text{ of }x=7x=\frac{81}{30}\text{ of }x=x=\frac{81}{30}\text{ of }x=2x=\frac{81}{30}\text{ of }x=27x=\frac{81}{30}\text{ of }x=\frac{27}{\placeholder{}}x=\frac{81}{30}\text{ of }x=27x=\frac{81}{30}\text{ of }x=2x=\frac{81}{30}\text{ of }x=2,
Controleer de oplossing door deze terug in de oorspronkelijke vergelijking te zetten.
Opdracht 2: Logaritmes samenvoegen
Opdracht: Los exact op:\log(4x+60)+\log(x)=3
Combineer de logaritmes: \log(\left(4x+60)\cdot x\right))=3\log(\left(4x+6)\cdot x\right))=3\log(4x+6)\cdot x)=3 Dit resulteert in: 4x^2+60x=10^3 (Hier geldt dat het grondtal 10 is, aangezien dit niet expliciet vermeld stond).
Dus,4x^2+60x=1000.4x^2+6x=1000.(4x^2+6x=1000.Herschrijf naar: 4x^2+60x-1000=0
Deel alles door 4 om eenvoudiger te kunnen werken: x^2+15x-250=0x^2+1x-250=0x^2+x-250=0x^2+\frac{}{2}x-250=0
Gebruik de som-product-methode: We willen twee getallen vinden die optellen tot1513\frac{3}{}en vermenigvuldigen tot-250.(-250.Geef de resultaten:
Oplossingen zijnenx=-25.(x=-25.Controleer welke voldoet; de negatieve waarde kan niet, omdat je geen negatieve argumenten in logaritmes kunt hebben.
Opdracht 3: Halve logaritmes
Opdracht: Los exact op:^{\frac12}\log(x-6)+4+^{\frac12}\log(x)=0^{\frac12}\log(x-6)+4+^{\frac{1}{\placeholder{}}}\log(x)=0^{\frac12}\log(x-6)+4+^1\log(x)=0^{\frac12}\log(x-6)+4+\log(x)=0^{\frac12}\log(x-6)+4+1\log(x)=0^{\frac12}\log(x-6)+4+\frac{1}{}\log(x)=0^{\frac12}\log(x-6)+4+\frac{1}{2}\log(x)=0^{\frac{1}{\placeholder{}}}\log(x-6)+4+\frac{1}{2}\log(x)=0^1\log(x-6)+4+\frac{1}{2}\log(x)=0\log(x-6)+4+\frac{1}{2}\log(x)=01\log(x-6)+4+\frac{1}{2}\log(x)=0\frac{1}{}\log(x-6)+4+\frac{1}{2}\log(x)=0
Brengnaar de andere kant: ^{\frac12}\log(x-6)+^{\frac12}\log(x)=-4^{\frac12}\log(x-6)+^{\frac{1}{\placeholder{}}}\log(x)=-4^{\frac12}\log(x-6)+^1\log(x)=-4^{\frac12}\log(x-6)+\log(x)=-4^{\frac12}\log(x-6)+1\log(x)=-4^{\frac12}\log(x-6)+\frac{1}{}\log(x)=-4^{\frac12}\log(x-6)+\frac{1}{2}\log(x)=-4^{\frac{1}{\placeholder{}}}\log(x-6)+\frac{1}{2}\log(x)=-4^1\log(x-6)+\frac{1}{2}\log(x)=-4\log(x-6)+\frac{1}{2}\log(x)=-4\frac{}{2}\log(x-6)+\frac{1}{2}\log(x)=-4
Combineer de logaritmen: ^{\frac12}\log(x^2-6x)=-4^{\frac12}\log(x^2!-6x)=-4^{\frac12}\log(x^2!\circ-6x)=-4^{\frac12}\log(x^2!\circ-6x)=-4^{\frac12}\log(x^2!\circ-6x)=-4^{\frac12}\log(x^2!-6x)=-4^{\frac12}\log(x^2!-6x)=-4^{\frac12}\log(x^2-6x)=-4^{\frac12}\log(x^2\circ-6x)=-4^{\frac12}\log(x^2-6x)=-4^{\frac12}\log(x^{}-6x)=-4^{\frac12}\log(x^{\circ}-6x)=-4^{\frac12}\log(x-6x)=-4^{\frac{1}{\placeholder{}}}\log(x-6x)=-4^1\log(x-6x)=-4\log(x-6x)=-42\log(x-6x)=-412\log(x-6x)=-4
Verwijder de halve logaritme door van beide zijden als exponent te schrijven: x^2-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{2^4}}=16x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{2^4}}=16x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{2^4}}=1x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{2^4}}=x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{2^4}}x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac12}x-6x=\left(\frac12\right)^{-4}=\frac{1}{\frac{1}{\placeholder{}}}x-6x=\left(\frac12\right)^{-4}=\frac11x-6x=\left(\frac12\right)^{-4}=\frac{1}{\placeholder{}}x-6x=\left(\frac12\right)^{-4}=1x-6x=\left(\frac12\right)^{-4}=x-6x=\left(\frac12\right)^{-4}x-6x=\left(\frac12\right)^{-}x-6x=\left(\frac12\right)^{-8}x-6x=\left(\frac12\right)^{-}x-6x=\left(\frac12\right)x-6x=\left(\frac12\right)x-6x=\frac12x-6x=\frac{1}{\placeholder{}}x-6x=1x-6x=x-6x=-x-6x=-8\log x-6x=-8\log(x-6x.=-8
Dit leidt tot een kwadratische vergelijking die je weer kan oplossen met de som-product-methode.
Probeer dit zelf! Als antwoord hoor jeente krijgen. Alleen de eerste oplossing voldoet, omdat je geen negatieve argumenten in logaritmes kan hebben.
Opdracht 4: Oplossen met minlogaritmes
Opdracht: Los exact op:^5\log\left(x-4_{}\right)+2=^5\log(3x)^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5^5\log\left(x-4_{}\right)+2=^5\log\left(x-4_{}\right)+2^5\log\left(x-4_{}\right)+^5\log\left(x-4_{}\right)^5\log\left(x-4_{}\right)=^5\log\left(x-4_{}\right)^5\log\left(x-4_{}\right)^5\log\left(x-4\right)^5\log\left(x-\right)^5\log\left(x\right)^5\log\left(\right)^5\log\left(\left(\right)\right)^5\log\left(\left(x\right)\right)^5\log\left(\left(x-\right)\right)^5\log\left(\left(x-4\right)\right)^5\log\left(\left(x-\right)\right)^5\log\left(\left(x\right)\right)^5\log\left(\left(\right)\right)^5\log\left(\right)^5\log\left(\right)^5\log^5^5^5^5^5
Breng5\log(3x)naar de andere kant: ^5\log\left(x-4\right)-^5\log\left(3x\right)=-2^5\log\left(x-4\right)-^5\log\left(3x\right)=-^5\log\left(x-4\right)-^5\log\left(3x\right)=^5\log\left(x-4\right)-^5\log\left(3x\right)=0^5\log\left(x-4\right)-^5\log\left(3x\right)=02^5\log\left(x-4\right)-^5\log\left(3x\right)=0^5\log\left(x-4\right)-^5\log\left(3x\right)=^5\log\left(x-4\right)-^5\log\left(3x\right)^5\log\left(x-4\right)-^5\log\left(3x\right)^5\log\left(x-4\right)-^5\log\left(3\right)^5\log\left(x-4\right)-^5\log\left(\right)^5\log\left(x-4\right)-^5\log^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5^5\log\left(x-4\right)-^5\log\left(x-4\right)^5\log\left(x-4\right)^5\log\left(x-\right)^5\log\left(x\right)^5\log\left(\right)^5\log^5^5^5^5^5
Gebruik de rekenregels om de logaritmes als één logaritme te schrijven:
^5\log\left(\frac{x-4}{3x}\right)=-2^5\log\left(\frac{x-4}{3x}\right)=-^5\log\left(\frac{x-4}{3x}\right)=^5\log\left(\frac{x-4}{3x}\right)=2^5\log\left(\frac{x-4}{3x}\right)=^5\log\left(\frac{x-4}{3x}\right)^5\log\left(\frac{x-4}{3x}\right)^5\log\left(\frac{x-4}{3}\right)^5\log\left(\frac{x-4}{\placeholder{}}\right)^5\log\left(\frac{x-}{\placeholder{}}\right)^5\log\left(\frac{x}{\placeholder{}}\right)^5\log\left(\frac{\placeholder{}}{\placeholder{}}\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log\left(\right)^5\log^5^5^5^5^5^5^5^5^{}^5^{\prime}^5^{\prime}\log^5^{\prime}\log^5^{\prime}lo^5^{\prime}l^5^{\prime}^5 Werkt de logaritme weg met behulp van exponenten:
\frac{x-4}{3x}=5^{-2}=\frac{1}{25}\frac{x-4}{3x}=5^{-2}=\frac12\frac{x-4}{3x}=5^{-2}=\frac{1}{\placeholder{}}\frac{x-4}{3x}=5^{-2}=1\frac{x-4}{3x}=5^{-2}=\frac{x-4}{3x}=5^{-2}\frac{x-4}{3x}=5^{-}\frac{x-4}{3x}=5\frac{x-4}{3x}=\frac{x-4}{3x}\frac{x-4}{3}\frac{x-4}{\placeholder{}}\frac{x-}{\placeholder{}}\frac{x}{\placeholder{}}\frac{}{\placeholder{}}\frac{\frac{\placeholder{}}{\placeholder{}}}{\placeholder{}}\frac{\frac{\placeholder{}}{\placeholder{}}}{\placeholder{}}\frac{fra}{\placeholder{}}\frac{fr}{\placeholder{}}\frac{f}{\placeholder{}}\frac{\placeholder{}}{\placeholder{}}
Doe beide kanten keerenom de breuken weg te werken:
25x-100=3x25x-10=3x25x-1=3x25x-1-=3x25x-1=3x25x-=3x25x-4=3x2x-4=3xx-4=3xx-4=\frac{3x}{}x-4=\frac{3x}{2}x-4=\frac{3x}{25}x-4=\frac{3x}{2}x-4=\frac{3x}{\placeholder{}}x-4=\frac{3}{\placeholder{}}x-4=\frac{\placeholder{}}{\placeholder{}}x-4=x-4x-x
Dit lost verder op tot:
22x=10022x=1022x=122x=1-22x=1--22x=1-22x=122x=22x222
x=\frac{100}{22}=4\frac{6}{11}x=\frac{100}{22}=\frac{6}{11}x=\frac{100}{22}=\frac61x=\frac{100}{22}=\frac{6}{\placeholder{}}x=\frac{100}{22}=6x=\frac{100}{22}=x=\frac{100}{22}x=\frac{100}{2}x=\frac{100}{\placeholder{}}x=100x=10x=1x=x
