Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 20 x^{1{,}09}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = 20 x^{1{,}09}\)
\(\log(y) = \log(20 x^{1{,}09})\)
1p
○
\(\log(y) = \log(20) + \log(x^{1{,}09})\)
\(\log(y) = \log(20) + 1{,}09 ⋅ \log(x)\)
1p
○
\(\log(y) = 1{,}301... + 1{,}09 ⋅ \log(x)\)
Dus \(y = 1{,}30 + 1{,}09 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {960 \over x^{4}}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {960 \over x^{4}} = 960 x^{-4}\)
\(\log(y) = \log(960 x^{-4})\)
1p
○
\(\log(y) = \log(960) + \log(x^{-4})\)
\(\log(y) = \log(960) - 4 ⋅ \log(x)\)
1p
○
\(\log(y) = 2{,}982... - 4 ⋅ \log(x)\)
Dus \(y = 2{,}98 - 4 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 3{,}54 + 1{,}83 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\)
Geef \(a\) in gehelen.
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\(\log(y) = 3{,}54 + 1{,}83 ⋅ \log(x)\)
\(\log(y) = \log(10^{3{,}54}) + \log(x^{1{,}83})\)
\(\log(y) = \log(10^{3{,}54} ⋅ x^{1{,}83})\)
1p
○
\(y = 10^{3{,}54} ⋅ x^{1{,}83}\)
1p
○
\(y = 3467{,}368... ⋅ x^{1{,}83}\)
Dus \(y = 3\,467 ⋅ x^{1{,}83} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 1\,100 ⋅ 1{,}08^{x}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 1\,100 ⋅ 1{,}08^{x}\)
\(\log(y) = \log(1\,100 ⋅ 1{,}08^{x})\)
\(\log(y) = \log(1\,100) + \log(1{,}08^{x})\)
1p
○
\(\log(y) = \log(1\,100) + x ⋅ \log(1{,}08)\)
1p
○
\(\log(y) = 3{,}041... + x ⋅ 0{,}03342...\)
Dus \(\log(y) = 0{,}0334 x + 3{,}04\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 2\,700 ⋅ 0{,}87^{4 x + 2}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 2\,700 ⋅ 0{,}87^{4 x + 2}\)
\(\log(y) = \log(2\,700 ⋅ 0{,}87^{4 x + 2})\)
\(\log(y) = \log(2\,700) + \log(0{,}87^{4 x + 2})\)
1p
○
\(\log(y) = \log(2\,700) + (4 x + 2) ⋅ \log(0{,}87)\)
\(\log(y) = \log(2\,700) + 4 x ⋅ \log(0{,}87) + 2 ⋅ \log(0{,}87)\)
1p
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\(\log(y) = 3{,}431... + 4 x ⋅ -0{,}06048... + 2 ⋅ -0{,}06048...\)
\(\log(y) = 3{,}431... - 0{,}24192... ⋅ x - 0{,}12096...\)
Dus \(\log(y) = -0{,}2419 x + 3{,}31\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = -0{,}1072 x + 1{,}95\) in de vorm \(y = b ⋅ g^{x} \text{.}\)
Geef \(b\) in gehelen en \(g\) in twee decimalen.
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\(\log(y) = -0{,}1072 x + 1{,}95\)
\(y = 10^{-0{,}1072 x + 1{,}95}\)
1p
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\(y = 10^{-0{,}1072 x} ⋅ 10^{1{,}95}\)
\(y = (10^{-0{,}1072})^{x} ⋅ 10^{1{,}95}\)
1p
○
\(y = 0{,}781...^{x} ⋅ 89{,}125...\)
Dus \(y = 89 ⋅ 0{,}78^{x} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 3{,}39 ⋅ {}^{3}\!\log(x) - 2{,}25\) in de vorm \(y = {}^{3}\!\log(a x^{b}) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
○
\(y = 3{,}39 ⋅ {}^{3}\!\log(x) - 2{,}25\)
\(\text{ } = {}^{3}\!\log(x^{3{,}39}) - 2{,}25\)
1p
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\(\text{ } = {}^{3}\!\log(x^{3{,}39}) + {}^{3}\!\log(3^{-2{,}25})\)
\(\text{ } = {}^{3}\!\log(x^{3{,}39} ⋅ 3^{-2{,}25})\)
1p
○
\(\text{ } = {}^{3}\!\log(x^{3{,}39} ⋅ 0{,}084...)\)
Dus \(y = {}^{3}\!\log(0{,}08 ⋅ x^{3{,}39}) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{2}\!\log(38 x^{3})\) in de vorm \(y = a + b ⋅ {}^{2}\!\log(x) \text{.}\)
Geef \(a\) in twee decimalen.
○
\(y = {}^{2}\!\log(38 x^{3})\)
\(\text{ } = {}^{2}\!\log(38 x^{3})\)
1p
○
\(\text{ } = {}^{2}\!\log(38) + {}^{2}\!\log(x^{3})\)
\(\text{ } = {}^{2}\!\log(38) + 3 ⋅ {}^{2}\!\log(x)\)
1p
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\(\text{ } = 5{,}247... + 3 ⋅ {}^{2}\!\log(x)\)
Dus \(y = 5{,}25 + 3 ⋅ {}^{2}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{4}\!\log(2{,}8 x) + 2{,}5\) in de vorm \(y = a + b ⋅ {}^{5}\!\log(x) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
○
\(y = {}^{4}\!\log(2{,}8 x) + 2{,}5\)
\(\text{ } = {}^{4}\!\log(2{,}8) + {}^{4}\!\log(x) + 2{,}5\)
1p
○
\(\text{ } = {}^{4}\!\log(2{,}8) + 2{,}5 + {{}^{5}\!\log(x) \over {}^{5}\!\log(4)}\)
\(\text{ } = {}^{4}\!\log(2{,}8) + 2{,}5 + {1 \over {}^{5}\!\log(4)} ⋅ {}^{5}\!\log(x)\)
1p
○
\(\text{ } = 0{,}742... + 2{,}5 + {1 \over 0{,}861...} ⋅ {}^{5}\!\log(x)\)
\(\text{ } = 3{,}242... + 1{,}160... ⋅ {}^{5}\!\log(x)\)
Dus \(y = 3{,}24 + 1{,}16 ⋅ {}^{5}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 8 ⋅ {}^{4}\!\log(128 x) + 10\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(2 x) \text{.}\)
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\(y = 8 ⋅ {}^{4}\!\log(128 x) + 10\)
\(\text{ } = 8 ⋅ ({}^{4}\!\log(64) + {}^{4}\!\log(2 x)) + 10\)
1p
○
\(\text{ } = 8 ⋅ (3 + {}^{4}\!\log(2 x)) + 10\)
1p
○
\(\text{ } = 24 + 8 ⋅ {}^{4}\!\log(2 x) + 10\)
\(\text{ } = 34 + 8 ⋅ {}^{4}\!\log(2 x)\)
1p
Druk \(x\) uit in \(y \text{.}\)
3p
\(y = 16 + 2 ⋅ {}^{4}\!\log(3 x + 9)\)
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\(y = 16 + 2 ⋅ {}^{4}\!\log(3 x + 9)\)
\(2 ⋅ {}^{4}\!\log(3 x + 9) = y - 16\)
\({}^{4}\!\log(3 x + 9) = \frac{1}{2} y - 8\)
1p
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\(3 x + 9 = 4^{\frac{1}{2} y - 8}\)
1p
○
\(3 x = 4^{\frac{1}{2} y - 8} - 9\)
\(x = \frac{1}{3} ⋅ 4^{\frac{1}{2} y - 8} - 3\)
1p