Getal & Ruimte (12e editie) - havo wiskunde B

'Logaritmische formules herleiden'.

havo wiskunde B	9.2 Werken met logaritmen
Logaritmische formules herleiden (1)	opgave 1 Druk \(x\) uit in \(y \text{.}\) 3p \(y = 18 + 3 ⋅ {}^{5}\!\log(2 x + 1)\) Vrijmaken 00kn - Logaritmische formules herleiden - basis - 0ms - dynamic variables ○ \(y = 18 + 3 ⋅ {}^{5}\!\log(2 x + 1)\) \(3 ⋅ {}^{5}\!\log(2 x + 1) = y - 18\) \({}^{5}\!\log(2 x + 1) = \frac{1}{3} y - 6\) 1p ○ \(2 x + 1 = 5^{\frac{1}{3} y - 6}\) 1p ○ \(2 x = 5^{\frac{1}{3} y - 6} - 1\) \(x = \frac{1}{2} ⋅ 5^{\frac{1}{3} y - 6} - \frac{1}{2}\) 1p
havo wiskunde B	9.3 Rekenregels voor logaritmen
Logaritmische formules herleiden (4)	opgave 1 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 1{,}45 ⋅ {}^{3}\!\log(x) - 2{,}52\) in de vorm \(y = {}^{3}\!\log(a x^{b}) \text{.}\) Geef \(a\) en \(b\) in twee decimalen. Herleiden (4) 00l0 - Logaritmische formules herleiden - basis - 0ms - dynamic variables a \(y = 1{,}45 ⋅ {}^{3}\!\log(x) - 2{,}52\) \(\text{ } = {}^{3}\!\log(x^{1{,}45}) - 2{,}52\) 1p ○ \(\text{ } = {}^{3}\!\log(x^{1{,}45}) + {}^{3}\!\log(3^{-2{,}52})\) \(\text{ } = {}^{3}\!\log(x^{1{,}45} ⋅ 3^{-2{,}52})\) 1p ○ \(\text{ } = {}^{3}\!\log(x^{1{,}45} ⋅ 0{,}062...)\) Dus \(y = {}^{3}\!\log(0{,}06 ⋅ x^{1{,}45}) \text{.}\) 1p 3p b Schrijf de formule \(y = {}^{2}\!\log({56 \over x^{3} \sqrt{x}})\) in de vorm \(y = a + b ⋅ {}^{2}\!\log(x) \text{.}\) Geef \(a\) in twee decimalen. Logaritmisch (5) 00l1 - Logaritmische formules herleiden - basis - 0ms - dynamic variables b \(y = {}^{2}\!\log({56 \over x^{3} \sqrt{x}})\) \(\text{ } = {}^{2}\!\log(56 x^{-3{,}5})\) 1p ○ \(\text{ } = {}^{2}\!\log(56) + {}^{2}\!\log(x^{-3{,}5})\) \(\text{ } = {}^{2}\!\log(56) - 3{,}5 ⋅ {}^{2}\!\log(x)\) 1p ○ \(\text{ } = 5{,}807... - 3{,}5 ⋅ {}^{2}\!\log(x)\) Dus \(y = 5{,}81 - 3{,}5 ⋅ {}^{2}\!\log(x) \text{.}\) 1p 3p c Schrijf de formule \(y = {}^{3}\!\log(1{,}4 x) + 1{,}4\) in de vorm \(y = a + b ⋅ {}^{5}\!\log(x) \text{.}\) Geef \(a\) en \(b\) in twee decimalen. Herleiden (6) 00l2 - Logaritmische formules herleiden - basis - 0ms - dynamic variables c \(y = {}^{3}\!\log(1{,}4 x) + 1{,}4\) \(\text{ } = {}^{3}\!\log(1{,}4) + {}^{3}\!\log(x) + 1{,}4\) 1p ○ \(\text{ } = {}^{3}\!\log(1{,}4) + 1{,}4 + {{}^{5}\!\log(x) \over {}^{5}\!\log(3)}\) \(\text{ } = {}^{3}\!\log(1{,}4) + 1{,}4 + {1 \over {}^{5}\!\log(3)} ⋅ {}^{5}\!\log(x)\) 1p ○ \(\text{ } = 0{,}306... + 1{,}4 + {1 \over 0{,}682...} ⋅ {}^{5}\!\log(x)\) \(\text{ } = 1{,}706... + 1{,}464... ⋅ {}^{5}\!\log(x)\) Dus \(y = 1{,}71 + 1{,}46 ⋅ {}^{5}\!\log(x) \text{.}\) 1p 3p d Schrijf de formule \(y = 9 ⋅ {}^{3}\!\log(162 x) - 7\) in de vorm \(y = a + b ⋅ {}^{3}\!\log(2 x) \text{.}\) Herleiden (7) 00l3 - Logaritmische formules herleiden - basis - 1ms - dynamic variables d \(y = 9 ⋅ {}^{3}\!\log(162 x) - 7\) \(\text{ } = 9 ⋅ ({}^{3}\!\log(81) + {}^{3}\!\log(2 x)) - 7\) 1p ○ \(\text{ } = 9 ⋅ (4 + {}^{3}\!\log(2 x)) - 7\) 1p ○ \(\text{ } = 36 + 9 ⋅ {}^{3}\!\log(2 x) - 7\) \(\text{ } = 29 + 9 ⋅ {}^{3}\!\log(2 x)\) 1p
havo wiskunde B	9.4 Formules omwerken
Logaritmische formules herleiden (6)	opgave 1 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 5\,600 ⋅ 0{,}77^{x}\) in de vorm \(\log(y) = a x + b \text{.}\) Geef \(a\) in vier decimalen en \(b\) in twee decimalen. Herleiden (1) 00ko - Logaritmische formules herleiden - basis - 0ms - dynamic variables a \(y = 5\,600 ⋅ 0{,}77^{x}\) \(\log(y) = \log(5\,600 ⋅ 0{,}77^{x})\) \(\log(y) = \log(5\,600) + \log(0{,}77^{x})\) 1p ○ \(\log(y) = \log(5\,600) + x ⋅ \log(0{,}77)\) 1p ○ \(\log(y) = 3{,}748... + x ⋅ -0{,}11350...\) Dus \(\log(y) = -0{,}1135 x + 3{,}75\) 1p 3p b Schrijf de formule \(y = 3\,200 ⋅ 1{,}1^{4 x + 3}\) in de vorm \(\log(y) = a x + b \text{.}\) Geef \(a\) in vier decimalen en \(b\) in twee decimalen. Herleiden (2) 00kp - Logaritmische formules herleiden - basis - 0ms - dynamic variables b \(y = 3\,200 ⋅ 1{,}1^{4 x + 3}\) \(\log(y) = \log(3\,200 ⋅ 1{,}1^{4 x + 3})\) \(\log(y) = \log(3\,200) + \log(1{,}1^{4 x + 3})\) 1p ○ \(\log(y) = \log(3\,200) + (4 x + 3) ⋅ \log(1{,}1)\) \(\log(y) = \log(3\,200) + 4 x ⋅ \log(1{,}1) + 3 ⋅ \log(1{,}1)\) 1p ○ \(\log(y) = 3{,}505... + 4 x ⋅ 0{,}04139... + 3 ⋅ 0{,}04139...\) \(\log(y) = 3{,}505... + 0{,}16557... ⋅ x + 0{,}12417...\) Dus \(\log(y) = 0{,}1656 x + 3{,}63\) 1p 3p c Schrijf de formule \(\log(y) = 0{,}9691 x + 3{,}13\) in de vorm \(y = b ⋅ g^{x} \text{.}\) Geef \(b\) in gehelen en \(g\) in twee decimalen. Herleiden (3) 00kq - Logaritmische formules herleiden - basis - 0ms - dynamic variables c \(\log(y) = 0{,}9691 x + 3{,}13\) \(y = 10^{0{,}9691 x + 3{,}13}\) 1p ○ \(y = 10^{0{,}9691 x} ⋅ 10^{3{,}13}\) \(y = (10^{0{,}9691})^{x} ⋅ 10^{3{,}13}\) 1p ○ \(y = 9{,}313...^{x} ⋅ 1348{,}962...\) Dus \(y = 1\,349 ⋅ 9{,}31^{x} \text{.}\) 1p 3p d Schrijf de formule \(\log(y) = 1{,}87 - 1{,}99 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\) Geef \(a\) in gehelen. Dubbel (3) 00kr - Logaritmische formules herleiden - basis - 0ms - dynamic variables d \(\log(y) = 1{,}87 - 1{,}99 ⋅ \log(x)\) \(\log(y) = \log(10^{1{,}87}) + \log(x^{-1{,}99})\) \(\log(y) = \log(10^{1{,}87} ⋅ x^{-1{,}99})\) 1p ○ \(y = 10^{1{,}87} ⋅ x^{-1{,}99}\) 1p ○ \(y = 74{,}131... ⋅ x^{-1{,}99}\) Dus \(y = 74 ⋅ x^{-1{,}99} \text{.}\) 1p opgave 2 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 80 x^{-1{,}76}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\) Geef \(a\) in twee decimalen. Dubbel (1) 00ks - Logaritmische formules herleiden - basis - 0ms - dynamic variables a \(y = 80 x^{-1{,}76}\) \(\log(y) = \log(80 x^{-1{,}76})\) 1p ○ \(\log(y) = \log(80) + \log(x^{-1{,}76})\) \(\log(y) = \log(80) - 1{,}76 ⋅ \log(x)\) 1p ○ \(\log(y) = 1{,}903... - 1{,}76 ⋅ \log(x)\) Dus \(y = 1{,}90 - 1{,}76 ⋅ \log(x) \text{.}\) 1p 3p b Schrijf de formule \(y = {10 \over \sqrt{x}}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\) Geef \(a\) in twee decimalen. Dubbel (2) 00kt - Logaritmische formules herleiden - basis - 0ms - dynamic variables b \(y = {10 \over \sqrt{x}} = 10 x^{-0{,}5}\) \(\log(y) = \log(10 x^{-0{,}5})\) 1p ○ \(\log(y) = \log(10) + \log(x^{-0{,}5})\) \(\log(y) = \log(10) - 0{,}5 ⋅ \log(x)\) 1p ○ \(\log(y) = 1 - 0{,}5 ⋅ \log(x)\) Dus \(y = 1{,}00 - 0{,}5 ⋅ \log(x) \text{.}\) 1p

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