Regels van logaritmen
Voordat we beginnen, is het handig om de basisregels van logaritmen nog eens op een rijtje te zetten:
Optelling van logaritmen: \log_{g}(a)+\log_{g}(b)=\log_{g}(ab)\log_{g}(a)+\log_{g}(b)=\log(ab)\log_{g}(a)+\log_{g}(b)=g\log(ab)\log_{g}(a)+\log(b)=g\log(ab)\log_{g}(a)+g\log(b)=g\log(ab)\log(a)+g\log(b)=g\log(ab)g\log(a) + g\log(b) = g\log(ab)
Aftrekken van logaritmen: \log_{g}(a)-\log_{g}(b)=\log_{g}\left(\frac{a}{b}\right)\log_{g}(a)-\log_{g}(b)=\log\left(\frac{a}{b}\right)\log_{g}(a)-\log_{g}(b)=g\log\left(\frac{a}{b}\right)\log_{g}(a)-\log(b)=g\log\left(\frac{a}{b}\right)\log_{g}(a)-g\log(b)=g\log\left(\frac{a}{b}\right)\log(a)-g\log(b)=g\log\left(\frac{a}{b}\right)
Exponent in logaritme: p\cdot\log_{g}(a)=\log_{g}(a^{p})p\cdot\log_{g}(a)=\log(a^{p})p\cdot\log_{g}(a)=g\log(a^{p})p\cdot\log(a)=g\log(a^{p})
Veranderen van grondtal: \log_{g}(a)=\frac{\log_{p}(a)}{\log_{p}(g)}\log_{g}(a)=\frac{\log_{p}(a)}{\log_{p}(g)}\log_{g}(a)=\frac{\log_{p}(a)}{\log(g)}\log_{g}(a)=\frac{\log_{p}(a)}{p\log(g)}\log_{g}(a)=\frac{\log(a)}{p\log(g)}\log_{g}(a)=\frac{p\log(a)}{p\log(g)}\log(a)=\frac{p\log(a)}{p\log(g)}
Rekenregels van machten
Daarnaast zijn er een paar belangrijke rekenregels van machten:
Product van machten:
Quotiënt van machten:
Macht van een macht:
Macht van nul:
Negatieve exponent:
Exponentiële formules omzetten
Voorbeeld 1
Gegeven: n=2400\cdot1{,}36^{t}n=2400\cdot136^{t}n=2400\cdot1m36^{t}n=2400\cdot136^{t}
We willen omzetten naar:
Waarinais afgerond op vier decimalen enop twee decimalen.
Neem de log aan beide kanten: \log(n)=\log(2400\cdot1{,}36^{t})\log(n)=\log(2400\cdot136^{t})\log(n)=\log(2400\cdot1.36^{t})\log(n)=\log(2400\cdot1.{,}36^{t})\log(n)=\log(2400\cdot1.{,}{,}36^{t})\log(n)=\log(2400\cdot1.{,}{,}]36^{t})\log(n)=\log(2400\cdot1.{,}{,}]]36^{t})\log(n)=\log(2400\cdot1.{,}{,}]36^{t})\log(n)=\log(2400\cdot1.{,}{,}36^{t})\log(n)=\log(2400\cdot1.{,}36^{t})\log(n)=\log(2400\cdot1.{,}36^{t})\log(n)=\log(2400\cdot1.{,}36^{t})\log(n) = \log(2400 \cdot 1.36^t)
Gebruik de regels van logaritmen: \log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(1{,}36)\log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(136)\log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(1.36)\log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(1.{,}36)\log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(1.36)\log(n)=\log(2400)+\log(1{,}36^{t})=\log(2400)+t\cdot\log(1.36)\log(n)=\log(2400)+\log(136^{t})=\log(2400)+t\cdot\log(1.36)\log(n)=\log(2400)+\log(1.36^{t})=\log(2400)+t\cdot\log(1.36)\log(n)=\log(2400)+\log(1.{,}36^{t})=\log(2400)+t\cdot\log(1.36) \log(n) = \log(2400) + \log(1.36^t) = \log(2400) + t\cdot\log(1.36)
Reken de logaritmes uit:
\log(2400)\approx3{,}38\log(2400)\approx338(2 decimalen)
\log(1{,}36)\approx0{,}1335\log(1{,}36)\approx01335\log(1{,}36)\approx0.1335\log(1{,}36)\approx0.133\log(1{,}36)\approx0.13\log(1{,}36)\approx0.134\log(1{,}36)\approx0.1345\log(1{,}36)\approx0.134\log(1{,}36)\approx0.13\log(1{,}36)\approx0.135\log(1{,}36)\approx0.13\log(1{,}36)\approx0.130\log(1{,}36)\approx0.1304\log(136)\approx0.1304 (4 decimalen)
Dus de formule wordt: \log(n)=0{,}1335t+3{,}38\log(n)=0{,}1335t+338\log(n)=0{,}1335t+3.38\log(n)=0{,}133t+3.38\log(n)=0{,}13t+3.38\log(n)=0{,}1t+3.38\log(n)=0{,}t+3.38\log(n)=0t+3.38\log(n)=0.t+3.38\log(n)=0.1t+3.38\log(n)=0.13t+3.38\log(n)=0.130t+3.38
Voorbeeld 2
Gegeven: n=13000\cdot0{,}56^{(t-4)}n=13000\cdot056^{(t-4)}
We willen dit schrijven in de vorm:
Waarinais afgerond op vier decimalen enop twee decimalen.
Neem de log aan beide kanten: \log(n)=\log(13000\cdot0{,}56^{(t-4)})\log(n)=\log(13000\cdot056^{(t-4)})
Splits de logaritmen: \log(n)=\log(13000)+\log(0{,}56^{(t-4)})=\log(13000)+(t-4)\cdot\log(0{,}56)\log(n)=\log(13000)+\log(0{,}56^{(t-4)})=\log(13000)+(t-4)\cdot\log(056)\log(n)=\log(13000)+\log(0{,}56^{(t-4)})=\log(13000)+(t-4)\cdot\log(0.56)\log(n)=\log(13000)+\log(056^{(t-4)})=\log(13000)+(t-4)\cdot\log(0.56)
Reken uit:
\log(13000)\approx4{,}11\log(13000)\approx4{,}113\log(13000)\approx4113\log(13000)\approx4.113(2 decimalen)
\log(0{,}56)\approx-0{,}2518\log(0{,}56)\approx-02518\log(0{,}56)\approx-0.2518\log(056)\approx-0.2518(4 decimalen)
Nadat je de haakjes hebt uitgewerkt, krijg je: \log(n)=-0{,}2518\left(t-4)+4{,}11\right.=-0{,}2518t+\left(4{,}11+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4{,}11\right.=-0{,}2518t+\left(4{,}113+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4{,}11\right.=-02518t+\left(4{,}113+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4{,}11\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4{,}113\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4113\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0{,}2518\left(t-4)+4.113\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-02518\left(t-4)+4.113\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0.2518\left(t-4)+4.113\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}113+1{,}007\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+1{,}007\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+1{,}00\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+1{,}0\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+1{,}\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+1\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139+\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1139\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}113\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}11\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}1\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4{,}\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(4\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\left(\right)\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+\log(n)=-0.2518\left(t-4)+4.1139\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1.\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1.0\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1.00\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1.007\right.=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+4.1139+1.007\right)=-0.2518t+5.12\log(n)=-0.2518\left(t-4)+(4.1139+1.007\right)=-0.2518t+5.12\log(n)=-0.2518\left(t-4+(4.1139+1.007\right)=-0.2518t+5.12\log(n)=-0.2518\left(t-+(4.1139+1.007\right)=-0.2518t+5.12\log(n)=-0.2518\left(t+(4.1139+1.007\right)=-0.2518t+5.12
Uiteindelijk krijg je dus:
\log\left(n\right)=5{,}12-0{,}2518t\log\left(n\right)=5{,}12+0{,}2518t\log\left(n\right)=5{,}12+0{,}2518\log\left(n\right)=5{,}12+0{,}251\log\left(n\right)=5{,}12+0{,}25\log\left(n\right)=5{,}12+0{,}2\log\left(n\right)=5{,}12+0{,}\log\left(n\right)=5{,}12+0\log\left(n\right)=5{,}12+\log\left(n\right)=5{,}12\log\left(n\right)=5{,}1\log\left(n\right)=5{,}\log\left(n\right)=5\log\left(n\right)=\log\left(n\right)\log\left(n\right)\log\left(\right)\log
Voorbeeld 3
Gegeven: n=5200\cdot1{,}65^{(2t-3)}n=5200\cdot165^{(2t-3)}n=5200\cdot1.65^{(2t-3)}n=5200\cdot1.{,}65^{(2t-3)}
We willen dit schrijven in de vorm:
Waarinais afgerond op vier decimalen enop twee decimalen.
Probeer deze opdracht zelf! Volg dezelfde stappen als bij de vorige opdrachten. Neem de log, splits op, en reken alles uit. Vergeet niet de terminologie correct te hanteren en de getallen goed af te ronden!
Als eindantwoord krijg je:
\log\left(n\right)=3{,}06+0{,}4350t\log\left(n\right)=3{,}06+0{,}4350\log\left(n\right)=3{,}06+0{,}435\log\left(n\right)=3{,}06+0{,}43\log\left(n\right)=3{,}06+0{,}4\log\left(n\right)=3{,}06+0{,}\log\left(n\right)=3{,}06+0\log\left(n\right)=3{,}06+\log\left(n\right)=3{,}06\log\left(n\right)=3{,}0\log\left(n\right)=3{,}\log\left(n\right)=3\log\left(n\right)=\log\left(n\right)\log\left(n\right)\log\left(\right)\log\frac{\placeholder{}}{\placeholder{}}
Machtsformules omzetten
Voorbeeld 1
Gegeven: y=20x^{1{,}6}y=20x^{16}y=20x^{1.6}=20x^{1.6}
We willen dit schrijven in de vorm:
\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(\right)\log\left(y\right)=a+b\cdot\log\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+\frac{b}{}\log\left(y\right)=a+\frac{b}{d}\log\left(y\right)=a+\frac{b}{\placeholder{}}\log\left(y\right)=a+b\log\left(y\right)=a+\log\left(y\right)=a\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(\right)\log
Geefin twee decimalen nauwkeurig.
Neem de logaritme aan beide kanten: \log(y)=\log(20)+\log(x^{1{,}6})\log(y)=\log(20)+\log(x^{16})\log(y)=\log(20)+\log(x^{1.6})\log()=\log(20)+\log(x^{1.6})\log(i)=\log(20)+\log(x^{1.6})\log(iy)=\log(20)+\log(x^{1.6})
Regels toepassen: \log(y)=\log(20)+1{,}6\cdot\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+16\log(x)\log(y)=\log(20)+1.6\log(x)\log()=\log(20)+1.6\log(x)
Rondaf op 2 decimalen:1{,}30.130.1.30.(1.30.
Uiteindelijk krijg je dan:
\log\left(y\right)=1{,}30+1{,}6\cdot\log\left(x\right)\log\left(y\right)=1{,}30+1{,}6\cdot\log\left(x\right)\log\left(y\right)=1{,}30+1{,}6\cdot\log\left(\right)\log\left(y\right)=1{,}30+1{,}6\cdot\log\log\left(y\right)=1{,}30+1{,}6\cdot lo\log\left(y\right)=1{,}30+1{,}6\cdot l\log\left(y\right)=1{,}30+1{,}6\cdot\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}6\log\left(y\right)=1{,}30+1{,}\log\left(y\right)=1{,}30+1\log\left(y\right)=1{,}30+\log\left(y\right)=1{,}30\log\left(y\right)=1{,}3\log\left(y\right)=1{,}\log\left(y\right)=1\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(\right)\log
Voorbeeld 2
Gegeven: y=78x^{-3{,}9}y=78x^{-39}y=78x^{-3.9}=78x^{-3.9}
We willen dit schrijven in de vorm:
\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(\right)\log\left(y\right)=a+b\cdot\log\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+\frac{b}{}\log\left(y\right)=a+\frac{b}{d}\log\left(y\right)=a+\frac{b}{\placeholder{}}\log\left(y\right)=a+b\log\left(y\right)=a+\log\left(y\right)=a\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(\right)\log
Geefin twee decimalen nauwkeurig.
Neem de log aan beide kanten: \log(y)=\log(78)+\log(x^{-3{,}9})\log(y)=\log(78)+\log(x^{-39})\log(y)=\log(78)+\log(x^{-3.9})\log()=\log(78)+\log(x^{-3.9})
Regels toepassen: \log(y)=\log(78)-3{,}9\cdot\log(x)\log(y)=\log(78)-3{,}\cdot\log(x)\log(y)=\log(78)-3\cdot\log(x)\log(y)=\log(78)-\cdot\log(x)\log(y)=\log(78)\cdot\log(x)\log(y)=\log(78)+\cdot\log(x)\log(y)=\log(78)+1\cdot\log(x)\log(y)=\log(78)+1{,}\cdot\log(x)\log(y)=\log(78)+1{,}6\cdot\log(x)\log(y)=\log(7)+1{,}6\cdot\log(x)\log(y)=\log()+1{,}6\cdot\log(x)\log(y)=\log(2)+1{,}6\cdot\log(x)\log(y)=\log(20)+1{,}6\cdot\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+1{,}6\log(x)\log(y)=\log(20)+16\log(x)\log(y)=\log(20)+1.6\log(x)\log()=\log(20)+1.6\log(x)
Reken uit: \log(78)\approx1{,}89.\log(78)\approx189.\log(78)\approx1.89.\log(78)\approx1.89).
Het eindantwoord is dan:
\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(x\right)\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(x\right)\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(x0\right)\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(x\right)\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(\right)\log\left(y\right)=1{,}89-3{,}9\cdot\log\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(y\right)=1{,}89-3{,}9\cdot\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}9\log\left(y\right)=1{,}89-3{,}\log\left(y\right)=1{,}89-3\log\left(y\right)=1{,}89-\log\left(y\right)=1{,}89\log\left(y\right)=1{,}89+\log\left(y\right)=1{,}89\log\left(y\right)=1{,}8\log\left(y\right)=1{,}\log\left(y\right)=1\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(\right)\log
Voorbeeld 3
Gegeven: y=\frac{320}{x^3\sqrt{x}}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3\sqrt{}}y=\frac{320}{x^3\sqrt{}x}y=\frac{320}{x^3\sqrt{}x}y=\frac{320}{x^3\sqrt{}}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^3}y=\frac{320}{x^{\frac{3}{}}}y=\frac{320}{x^{\frac{3}{2}}}=\frac{320}{x^{\frac{3}{2}}}
We willen dit schrijven in de vorm:
\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(x\right)\log\left(y\right)=a+b\cdot\log\left(\right)\log\left(y\right)=a+b\cdot\log\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\cdot\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+b\log\left(y\right)=a+\frac{b}{}\log\left(y\right)=a+\frac{b}{d}\log\left(y\right)=a+\frac{b}{\placeholder{}}\log\left(y\right)=a+b\log\left(y\right)=a+\log\left(y\right)=a\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(\right)\log
Geefin twee decimalen nauwkeurig.
Probeer deze opdracht zelf! Maak gebruik van de rekenregels bovenaan de samenvatting en kijk als dit nodig is terug naar de vorige voorbeelden.
Uiteindelijk krijg je:
\log\left(y\right)=2{,}51-3\frac12\log\left(x\right)\log\left(y\right)=2{,}51-3\frac12\log\left(x\right)\log\left(y\right)=2{,}51-3\frac12\log\left(x\right)\log\left(y\right)=2{,}51-3\frac12\log\left(x\right)\log\left(y\right)=2{,}51-3\frac12\log\left(\right)\log\left(y\right)=2{,}51-3\frac12\log\log\left(y\right)=2{,}51-3\frac12\log\left(y\right)=2{,}51-3\frac12\log\left(y\right)=2{,}51-3\frac12\log\left(y\right)=2{,}51-3\frac12\log\left(y\right)=2{,}51-3\frac12\log\left(y\right)=2{,}51-3\frac{1}{\placeholder{}}\log\left(y\right)=2{,}51-3\frac{\placeholder{}}{\placeholder{}}\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-3\log\left(y\right)=2{,}51-\log\left(y\right)=2{,}51\log\left(y\right)=2{,}5\log\left(y\right)=2{,}\log\left(y\right)=2\log\left(y\right)=\log\left(y\right)\log\left(y\right)\log\left(y\right)\log\left(\right)\log\left(t\right)\log\left(\right)\log
