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 \(y = 32 + 4 ⋅ {}^{3}\!\log(6 x + 7)\) Vrijmaken 00kn - Logaritmische formules herleiden - basis - 1ms - dynamic variables ○ \(y = 32 + 4 ⋅ {}^{3}\!\log(6 x + 7)\) \(4 ⋅ {}^{3}\!\log(6 x + 7) = y - 32\) \({}^{3}\!\log(6 x + 7) = \frac{1}{4} y - 8\) 1p ○ \(6 x + 7 = 3^{\frac{1}{4} y - 8}\) 1p ○ \(6 x = 3^{\frac{1}{4} y - 8} - 7\) \(x = \frac{1}{6} ⋅ 3^{\frac{1}{4} y - 8} - 1\frac{1}{6}\) 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 \(y = 1{,}26 ⋅ {}^{2}\!\log(x) + 1{,}36\) in de vorm \(y = {}^{2}\!\log(a x^{b}) \text{.}\) Geef \(a\) en \(b\) in twee decimalen. Herleiden (4) 00l0 - Logaritmische formules herleiden - basis - 0ms - dynamic variables a \(y = 1{,}26 ⋅ {}^{2}\!\log(x) + 1{,}36\) \(\text{ } = {}^{2}\!\log(x^{1{,}26}) + 1{,}36\) 1p ○ \(\text{ } = {}^{2}\!\log(x^{1{,}26}) + {}^{2}\!\log(2^{1{,}36})\) \(\text{ } = {}^{2}\!\log(x^{1{,}26} ⋅ 2^{1{,}36})\) 1p ○ \(\text{ } = {}^{2}\!\log(x^{1{,}26} ⋅ 2{,}566...)\) Dus \(y = {}^{2}\!\log(2{,}57 ⋅ x^{1{,}26}) \text{.}\) 1p 3p b Schrijf de formule \(y = {}^{2}\!\log({47 \over x^{3} \sqrt{x}})\) 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 \(y = {}^{2}\!\log({47 \over x^{3} \sqrt{x}})\) \(\text{ } = {}^{2}\!\log(47 x^{-3{,}5})\) 1p ○ \(\text{ } = {}^{2}\!\log(47) + {}^{2}\!\log(x^{-3{,}5})\) \(\text{ } = {}^{2}\!\log(47) - 3{,}5 ⋅ {}^{2}\!\log(x)\) 1p ○ \(\text{ } = 5{,}554... - 3{,}5 ⋅ {}^{2}\!\log(x)\) Dus \(y = 5{,}55 - 3{,}5 ⋅ {}^{2}\!\log(x) \text{.}\) 1p 3p c Schrijf de formule \(y = {}^{3}\!\log(1{,}3 x) - 2{,}1\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(x) \text{.}\) Geef \(a\) en \(b\) in twee decimalen. Herleiden (6) 00l2 - Logaritmische formules herleiden - basis - 0ms - dynamic variables c \(y = {}^{3}\!\log(1{,}3 x) - 2{,}1\) \(\text{ } = {}^{3}\!\log(1{,}3) + {}^{3}\!\log(x) - 2{,}1\) 1p ○ \(\text{ } = {}^{3}\!\log(1{,}3) - 2{,}1 + {{}^{4}\!\log(x) \over {}^{4}\!\log(3)}\) \(\text{ } = {}^{3}\!\log(1{,}3) - 2{,}1 + {1 \over {}^{4}\!\log(3)} ⋅ {}^{4}\!\log(x)\) 1p ○ \(\text{ } = 0{,}238... - 2{,}1 + {1 \over 0{,}792...} ⋅ {}^{4}\!\log(x)\) \(\text{ } = -1{,}861... + 1{,}261... ⋅ {}^{4}\!\log(x)\) Dus \(y = -1{,}86 + 1{,}26 ⋅ {}^{4}\!\log(x) \text{.}\) 1p 3p d Schrijf de formule \(y = 5 ⋅ {}^{2}\!\log(48 x) - 10\) in de vorm \(y = a + b ⋅ {}^{2}\!\log(3 x) \text{.}\) Herleiden (7) 00l3 - Logaritmische formules herleiden - basis - 1ms - dynamic variables d \(y = 5 ⋅ {}^{2}\!\log(48 x) - 10\) \(\text{ } = 5 ⋅ ({}^{2}\!\log(16) + {}^{2}\!\log(3 x)) - 10\) 1p ○ \(\text{ } = 5 ⋅ (4 + {}^{2}\!\log(3 x)) - 10\) 1p ○ \(\text{ } = 20 + 5 ⋅ {}^{2}\!\log(3 x) - 10\) \(\text{ } = 10 + 5 ⋅ {}^{2}\!\log(3 x)\) 1p
havo wiskunde B	9.4 Formules omwerken
Logaritmische formules herleiden (6)	opgave 1 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 2\,300 ⋅ 0{,}79^{x}\) 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 \(y = 2\,300 ⋅ 0{,}79^{x}\) \(\log(y) = \log(2\,300 ⋅ 0{,}79^{x})\) \(\log(y) = \log(2\,300) + \log(0{,}79^{x})\) 1p ○ \(\log(y) = \log(2\,300) + x ⋅ \log(0{,}79)\) 1p ○ \(\log(y) = 3{,}361... + x ⋅ -0{,}10237...\) Dus \(\log(y) = -0{,}1024 x + 3{,}36\) 1p 3p b Schrijf de formule \(y = 1\,000 ⋅ 1{,}1^{5 x + 1}\) 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 - 1ms - dynamic variables b \(y = 1\,000 ⋅ 1{,}1^{5 x + 1}\) \(\log(y) = \log(1\,000 ⋅ 1{,}1^{5 x + 1})\) \(\log(y) = \log(1\,000) + \log(1{,}1^{5 x + 1})\) 1p ○ \(\log(y) = \log(1\,000) + (5 x + 1) ⋅ \log(1{,}1)\) \(\log(y) = \log(1\,000) + 5 x ⋅ \log(1{,}1) + 1 ⋅ \log(1{,}1)\) 1p ○ \(\log(y) = 3 + 5 x ⋅ 0{,}04139... + 1 ⋅ 0{,}04139...\) \(\log(y) = 3 + 0{,}20696... ⋅ x + 0{,}04139...\) Dus \(\log(y) = 0{,}2070 x + 3{,}04\) 1p 3p c Schrijf de formule \(\log(y) = -0{,}1217 x + 1{,}71\) 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 \(\log(y) = -0{,}1217 x + 1{,}71\) \(y = 10^{-0{,}1217 x + 1{,}71}\) 1p ○ \(y = 10^{-0{,}1217 x} ⋅ 10^{1{,}71}\) \(y = (10^{-0{,}1217})^{x} ⋅ 10^{1{,}71}\) 1p ○ \(y = 0{,}755...^{x} ⋅ 51{,}286...\) Dus \(y = 51 ⋅ 0{,}76^{x} \text{.}\) 1p 3p d Schrijf de formule \(\log(y) = 3{,}03 + 1{,}47 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\) Geef \(a\) in gehelen. Dubbel (3) 00kr - Logaritmische formules herleiden - basis - 0ms - dynamic variables d \(\log(y) = 3{,}03 + 1{,}47 ⋅ \log(x)\) \(\log(y) = \log(10^{3{,}03}) + \log(x^{1{,}47})\) \(\log(y) = \log(10^{3{,}03} ⋅ x^{1{,}47})\) 1p ○ \(y = 10^{3{,}03} ⋅ x^{1{,}47}\) 1p ○ \(y = 1071{,}519... ⋅ x^{1{,}47}\) Dus \(y = 1\,072 ⋅ x^{1{,}47} \text{.}\) 1p opgave 2 Herleid tot de gevraagde vorm. 3p a Schrijf de formule \(y = 120 x^{1{,}88}\) 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 \(y = 120 x^{1{,}88}\) \(\log(y) = \log(120 x^{1{,}88})\) 1p ○ \(\log(y) = \log(120) + \log(x^{1{,}88})\) \(\log(y) = \log(120) + 1{,}88 ⋅ \log(x)\) 1p ○ \(\log(y) = 2{,}079... + 1{,}88 ⋅ \log(x)\) Dus \(y = 2{,}08 + 1{,}88 ⋅ \log(x) \text{.}\) 1p 3p b Schrijf de formule \(y = {50 \over x^{3} \sqrt{x}}\) 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 \(y = {50 \over x^{3} \sqrt{x}} = 50 x^{-3{,}5}\) \(\log(y) = \log(50 x^{-3{,}5})\) 1p ○ \(\log(y) = \log(50) + \log(x^{-3{,}5})\) \(\log(y) = \log(50) - 3{,}5 ⋅ \log(x)\) 1p ○ \(\log(y) = 1{,}698... - 3{,}5 ⋅ \log(x)\) Dus \(y = 1{,}70 - 3{,}5 ⋅ \log(x) \text{.}\) 1p

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