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 \(y = 15 + 3 ⋅ {}^{2}\!\log(4 x + 8)\) Vrijmaken 00kn - Logaritmische formules herleiden - basis - 0ms - dynamic variables ○ \(y = 15 + 3 ⋅ {}^{2}\!\log(4 x + 8)\) \(3 ⋅ {}^{2}\!\log(4 x + 8) = y - 15\) \({}^{2}\!\log(4 x + 8) = \frac{1}{3} y - 5\) 1p ○ \(4 x + 8 = 2^{\frac{1}{3} y - 5}\) 1p ○ \(4 x = 2^{\frac{1}{3} y - 5} - 8\) \(x = \frac{1}{4} ⋅ 2^{\frac{1}{3} y - 5} - 2\) 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 \(y = 2{,}48 ⋅ {}^{3}\!\log(x) - 2{,}27\) 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 \(y = 2{,}48 ⋅ {}^{3}\!\log(x) - 2{,}27\) \(\text{ } = {}^{3}\!\log(x^{2{,}48}) - 2{,}27\) 1p ○ \(\text{ } = {}^{3}\!\log(x^{2{,}48}) + {}^{3}\!\log(3^{-2{,}27})\) \(\text{ } = {}^{3}\!\log(x^{2{,}48} ⋅ 3^{-2{,}27})\) 1p ○ \(\text{ } = {}^{3}\!\log(x^{2{,}48} ⋅ 0{,}082...)\) Dus \(y = {}^{3}\!\log(0{,}08 ⋅ x^{2{,}48}) \text{.}\) 1p 3p b Schrijf de formule \(y = {}^{3}\!\log({75 \over x^{4} \sqrt{x}})\) in de vorm \(y = a + b ⋅ {}^{3}\!\log(x) \text{.}\) Geef \(a\) in twee decimalen. Logaritmisch (5) 00l1 - Logaritmische formules herleiden - basis - 0ms - dynamic variables b \(y = {}^{3}\!\log({75 \over x^{4} \sqrt{x}})\) \(\text{ } = {}^{3}\!\log(75 x^{-4{,}5})\) 1p ○ \(\text{ } = {}^{3}\!\log(75) + {}^{3}\!\log(x^{-4{,}5})\) \(\text{ } = {}^{3}\!\log(75) - 4{,}5 ⋅ {}^{3}\!\log(x)\) 1p ○ \(\text{ } = 3{,}929... - 4{,}5 ⋅ {}^{3}\!\log(x)\) Dus \(y = 3{,}93 - 4{,}5 ⋅ {}^{3}\!\log(x) \text{.}\) 1p 3p c Schrijf de formule \(y = {}^{3}\!\log(2{,}3 x) - 0{,}9\) 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 \(y = {}^{3}\!\log(2{,}3 x) - 0{,}9\) \(\text{ } = {}^{3}\!\log(2{,}3) + {}^{3}\!\log(x) - 0{,}9\) 1p ○ \(\text{ } = {}^{3}\!\log(2{,}3) - 0{,}9 + {{}^{5}\!\log(x) \over {}^{5}\!\log(3)}\) \(\text{ } = {}^{3}\!\log(2{,}3) - 0{,}9 + {1 \over {}^{5}\!\log(3)} ⋅ {}^{5}\!\log(x)\) 1p ○ \(\text{ } = 0{,}758... - 0{,}9 + {1 \over 0{,}682...} ⋅ {}^{5}\!\log(x)\) \(\text{ } = -0{,}141... + 1{,}464... ⋅ {}^{5}\!\log(x)\) Dus \(y = -0{,}14 + 1{,}46 ⋅ {}^{5}\!\log(x) \text{.}\) 1p 3p d Schrijf de formule \(y = 8 ⋅ {}^{4}\!\log(128 x) + 5\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(2 x) \text{.}\) Herleiden (7) 00l3 - Logaritmische formules herleiden - basis - 1ms - dynamic variables d \(y = 8 ⋅ {}^{4}\!\log(128 x) + 5\) \(\text{ } = 8 ⋅ ({}^{4}\!\log(64) + {}^{4}\!\log(2 x)) + 5\) 1p ○ \(\text{ } = 8 ⋅ (3 + {}^{4}\!\log(2 x)) + 5\) 1p ○ \(\text{ } = 24 + 8 ⋅ {}^{4}\!\log(2 x) + 5\) \(\text{ } = 29 + 8 ⋅ {}^{4}\!\log(2 x)\) 1p
havo wiskunde B	9.4 Formules omwerken
Logaritmische formules herleiden (6)	opgave 1 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 5\,500 ⋅ 1{,}22^{x}\) 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 \(y = 5\,500 ⋅ 1{,}22^{x}\) \(\log(y) = \log(5\,500 ⋅ 1{,}22^{x})\) \(\log(y) = \log(5\,500) + \log(1{,}22^{x})\) 1p ○ \(\log(y) = \log(5\,500) + x ⋅ \log(1{,}22)\) 1p ○ \(\log(y) = 3{,}740... + x ⋅ 0{,}08635...\) Dus \(\log(y) = 0{,}0864 x + 3{,}74\) 1p 3p b Schrijf de formule \(y = 1\,500 ⋅ 1{,}06^{4 x + 6}\) 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 \(y = 1\,500 ⋅ 1{,}06^{4 x + 6}\) \(\log(y) = \log(1\,500 ⋅ 1{,}06^{4 x + 6})\) \(\log(y) = \log(1\,500) + \log(1{,}06^{4 x + 6})\) 1p ○ \(\log(y) = \log(1\,500) + (4 x + 6) ⋅ \log(1{,}06)\) \(\log(y) = \log(1\,500) + 4 x ⋅ \log(1{,}06) + 6 ⋅ \log(1{,}06)\) 1p ○ \(\log(y) = 3{,}176... + 4 x ⋅ 0{,}02530... + 6 ⋅ 0{,}02530...\) \(\log(y) = 3{,}176... + 0{,}10122... ⋅ x + 0{,}15183...\) Dus \(\log(y) = 0{,}1012 x + 3{,}33\) 1p 3p c Schrijf de formule \(\log(y) = 0{,}0077 x + 3{,}08\) 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 \(\log(y) = 0{,}0077 x + 3{,}08\) \(y = 10^{0{,}0077 x + 3{,}08}\) 1p ○ \(y = 10^{0{,}0077 x} ⋅ 10^{3{,}08}\) \(y = (10^{0{,}0077})^{x} ⋅ 10^{3{,}08}\) 1p ○ \(y = 1{,}017...^{x} ⋅ 1202{,}264...\) Dus \(y = 1\,202 ⋅ 1{,}02^{x} \text{.}\) 1p 3p d Schrijf de formule \(\log(y) = 2{,}14 - 1{,}29 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\) Geef \(a\) in gehelen. Dubbel (3) 00kr - Logaritmische formules herleiden - basis - 0ms - dynamic variables d \(\log(y) = 2{,}14 - 1{,}29 ⋅ \log(x)\) \(\log(y) = \log(10^{2{,}14}) + \log(x^{-1{,}29})\) \(\log(y) = \log(10^{2{,}14} ⋅ x^{-1{,}29})\) 1p ○ \(y = 10^{2{,}14} ⋅ x^{-1{,}29}\) 1p ○ \(y = 138{,}038... ⋅ x^{-1{,}29}\) Dus \(y = 138 ⋅ x^{-1{,}29} \text{.}\) 1p opgave 2 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 130 x^{-1{,}77}\) 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 \(y = 130 x^{-1{,}77}\) \(\log(y) = \log(130 x^{-1{,}77})\) 1p ○ \(\log(y) = \log(130) + \log(x^{-1{,}77})\) \(\log(y) = \log(130) - 1{,}77 ⋅ \log(x)\) 1p ○ \(\log(y) = 2{,}113... - 1{,}77 ⋅ \log(x)\) Dus \(y = 2{,}11 - 1{,}77 ⋅ \log(x) \text{.}\) 1p 3p b Schrijf de formule \(y = {760 \over x \sqrt{x}}\) 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 \(y = {760 \over x \sqrt{x}} = 760 x^{-1{,}5}\) \(\log(y) = \log(760 x^{-1{,}5})\) 1p ○ \(\log(y) = \log(760) + \log(x^{-1{,}5})\) \(\log(y) = \log(760) - 1{,}5 ⋅ \log(x)\) 1p ○ \(\log(y) = 2{,}880... - 1{,}5 ⋅ \log(x)\) Dus \(y = 2{,}88 - 1{,}5 ⋅ \log(x) \text{.}\) 1p

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