Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 110 x^{1{,}13}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = 110 x^{1{,}13}\)
\(\log(y) = \log(110 x^{1{,}13})\)
1p
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\(\log(y) = \log(110) + \log(x^{1{,}13})\)
\(\log(y) = \log(110) + 1{,}13 ⋅ \log(x)\)
1p
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\(\log(y) = 2{,}041... + 1{,}13 ⋅ \log(x)\)
Dus \(y = 2{,}04 + 1{,}13 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {390 \over x^{2} \sqrt{x}}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {390 \over x^{2} \sqrt{x}} = 390 x^{-2{,}5}\)
\(\log(y) = \log(390 x^{-2{,}5})\)
1p
○
\(\log(y) = \log(390) + \log(x^{-2{,}5})\)
\(\log(y) = \log(390) - 2{,}5 ⋅ \log(x)\)
1p
○
\(\log(y) = 2{,}591... - 2{,}5 ⋅ \log(x)\)
Dus \(y = 2{,}59 - 2{,}5 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 1{,}79 - 1{,}49 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\)
Geef \(a\) in gehelen.
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\(\log(y) = 1{,}79 - 1{,}49 ⋅ \log(x)\)
\(\log(y) = \log(10^{1{,}79}) + \log(x^{-1{,}49})\)
\(\log(y) = \log(10^{1{,}79} ⋅ x^{-1{,}49})\)
1p
○
\(y = 10^{1{,}79} ⋅ x^{-1{,}49}\)
1p
○
\(y = 61{,}659... ⋅ x^{-1{,}49}\)
Dus \(y = 62 ⋅ x^{-1{,}49} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 4\,300 ⋅ 1{,}3^{x}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 4\,300 ⋅ 1{,}3^{x}\)
\(\log(y) = \log(4\,300 ⋅ 1{,}3^{x})\)
\(\log(y) = \log(4\,300) + \log(1{,}3^{x})\)
1p
○
\(\log(y) = \log(4\,300) + x ⋅ \log(1{,}3)\)
1p
○
\(\log(y) = 3{,}633... + x ⋅ 0{,}11394...\)
Dus \(\log(y) = 0{,}1139 x + 3{,}63\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 9\,200 ⋅ 0{,}91^{2 x + 5}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 9\,200 ⋅ 0{,}91^{2 x + 5}\)
\(\log(y) = \log(9\,200 ⋅ 0{,}91^{2 x + 5})\)
\(\log(y) = \log(9\,200) + \log(0{,}91^{2 x + 5})\)
1p
○
\(\log(y) = \log(9\,200) + (2 x + 5) ⋅ \log(0{,}91)\)
\(\log(y) = \log(9\,200) + 2 x ⋅ \log(0{,}91) + 5 ⋅ \log(0{,}91)\)
1p
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\(\log(y) = 3{,}963... + 2 x ⋅ -0{,}04095... + 5 ⋅ -0{,}04095...\)
\(\log(y) = 3{,}963... - 0{,}08191... ⋅ x - 0{,}20479...\)
Dus \(\log(y) = -0{,}0819 x + 3{,}76\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 0{,}9134 x + 2{,}01\) in de vorm \(y = b ⋅ g^{x} \text{.}\)
Geef \(b\) in gehelen en \(g\) in twee decimalen.
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\(\log(y) = 0{,}9134 x + 2{,}01\)
\(y = 10^{0{,}9134 x + 2{,}01}\)
1p
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\(y = 10^{0{,}9134 x} ⋅ 10^{2{,}01}\)
\(y = (10^{0{,}9134})^{x} ⋅ 10^{2{,}01}\)
1p
○
\(y = 8{,}192...^{x} ⋅ 102{,}329...\)
Dus \(y = 102 ⋅ 8{,}19^{x} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 1{,}92 ⋅ {}^{4}\!\log(x) - 1{,}46\) in de vorm \(y = {}^{4}\!\log(a x^{b}) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
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\(y = 1{,}92 ⋅ {}^{4}\!\log(x) - 1{,}46\)
\(\text{ } = {}^{4}\!\log(x^{1{,}92}) - 1{,}46\)
1p
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\(\text{ } = {}^{4}\!\log(x^{1{,}92}) + {}^{4}\!\log(4^{-1{,}46})\)
\(\text{ } = {}^{4}\!\log(x^{1{,}92} ⋅ 4^{-1{,}46})\)
1p
○
\(\text{ } = {}^{4}\!\log(x^{1{,}92} ⋅ 0{,}132...)\)
Dus \(y = {}^{4}\!\log(0{,}13 ⋅ x^{1{,}92}) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{4}\!\log({47 \over x^{3} \sqrt{x}})\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(x) \text{.}\)
Geef \(a\) in twee decimalen.
○
\(y = {}^{4}\!\log({47 \over x^{3} \sqrt{x}})\)
\(\text{ } = {}^{4}\!\log(47 x^{-3{,}5})\)
1p
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\(\text{ } = {}^{4}\!\log(47) + {}^{4}\!\log(x^{-3{,}5})\)
\(\text{ } = {}^{4}\!\log(47) - 3{,}5 ⋅ {}^{4}\!\log(x)\)
1p
○
\(\text{ } = 2{,}777... - 3{,}5 ⋅ {}^{4}\!\log(x)\)
Dus \(y = 2{,}78 - 3{,}5 ⋅ {}^{4}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{2}\!\log(1{,}2 x) + 0{,}1\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(x) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
○
\(y = {}^{2}\!\log(1{,}2 x) + 0{,}1\)
\(\text{ } = {}^{2}\!\log(1{,}2) + {}^{2}\!\log(x) + 0{,}1\)
1p
○
\(\text{ } = {}^{2}\!\log(1{,}2) + 0{,}1 + {{}^{4}\!\log(x) \over {}^{4}\!\log(2)}\)
\(\text{ } = {}^{2}\!\log(1{,}2) + 0{,}1 + {1 \over {}^{4}\!\log(2)} ⋅ {}^{4}\!\log(x)\)
1p
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\(\text{ } = 0{,}263... + 0{,}1 + {1 \over 0{,}5} ⋅ {}^{4}\!\log(x)\)
\(\text{ } = 0{,}363... + 2 ⋅ {}^{4}\!\log(x)\)
Dus \(y = 0{,}36 + 2{,}00 ⋅ {}^{4}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 9 ⋅ {}^{4}\!\log(128 x) - 8\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(2 x) \text{.}\)
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\(y = 9 ⋅ {}^{4}\!\log(128 x) - 8\)
\(\text{ } = 9 ⋅ ({}^{4}\!\log(64) + {}^{4}\!\log(2 x)) - 8\)
1p
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\(\text{ } = 9 ⋅ (3 + {}^{4}\!\log(2 x)) - 8\)
1p
○
\(\text{ } = 27 + 9 ⋅ {}^{4}\!\log(2 x) - 8\)
\(\text{ } = 19 + 9 ⋅ {}^{4}\!\log(2 x)\)
1p
Druk \(x\) uit in \(y \text{.}\)
3p
\(y = 12 + 3 ⋅ {}^{2}\!\log(7 x + 1)\)
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\(y = 12 + 3 ⋅ {}^{2}\!\log(7 x + 1)\)
\(3 ⋅ {}^{2}\!\log(7 x + 1) = y - 12\)
\({}^{2}\!\log(7 x + 1) = \frac{1}{3} y - 4\)
1p
○
\(7 x + 1 = 2^{\frac{1}{3} y - 4}\)
1p
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\(7 x = 2^{\frac{1}{3} y - 4} - 1\)
\(x = \frac{1}{7} ⋅ 2^{\frac{1}{3} y - 4} - \frac{1}{7}\)
1p