Logaritmische formules herleiden
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 890 x^{1{,}43}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = 890 x^{1{,}43}\)
\(\log(y) = \log(890 x^{1{,}43})\)
1p
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\(\log(y) = \log(890) + \log(x^{1{,}43})\)
\(\log(y) = \log(890) + 1{,}43 ⋅ \log(x)\)
1p
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\(\log(y) = 2{,}949... + 1{,}43 ⋅ \log(x)\)
Dus \(y = 2{,}95 + 1{,}43 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {90 \over x^{3}}\) in de vorm \(\log(y) = a + b ⋅ \log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {90 \over x^{3}} = 90 x^{-3}\)
\(\log(y) = \log(90 x^{-3})\)
1p
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\(\log(y) = \log(90) + \log(x^{-3})\)
\(\log(y) = \log(90) - 3 ⋅ \log(x)\)
1p
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\(\log(y) = 1{,}954... - 3 ⋅ \log(x)\)
Dus \(y = 1{,}95 - 3 ⋅ \log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 1{,}79 - 1{,}79 ⋅ \log(x)\) in de vorm \(y = a x^{b} \text{.}\)
Geef \(a\) in gehelen.
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\(\log(y) = 1{,}79 - 1{,}79 ⋅ \log(x)\)
\(\log(y) = \log(10^{1{,}79}) + \log(x^{-1{,}79})\)
\(\log(y) = \log(10^{1{,}79} ⋅ x^{-1{,}79})\)
1p
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\(y = 10^{1{,}79} ⋅ x^{-1{,}79}\)
1p
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\(y = 61{,}659... ⋅ x^{-1{,}79}\)
Dus \(y = 62 ⋅ x^{-1{,}79} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 9\,800 ⋅ 1{,}17^{x}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 9\,800 ⋅ 1{,}17^{x}\)
\(\log(y) = \log(9\,800 ⋅ 1{,}17^{x})\)
\(\log(y) = \log(9\,800) + \log(1{,}17^{x})\)
1p
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\(\log(y) = \log(9\,800) + x ⋅ \log(1{,}17)\)
1p
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\(\log(y) = 3{,}991... + x ⋅ 0{,}06818...\)
Dus \(\log(y) = 0{,}0682 x + 3{,}99\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 9\,400 ⋅ 0{,}75^{5 x + 3}\) in de vorm \(\log(y) = a x + b \text{.}\)
Geef \(a\) in vier decimalen en \(b\) in twee decimalen.
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\(y = 9\,400 ⋅ 0{,}75^{5 x + 3}\)
\(\log(y) = \log(9\,400 ⋅ 0{,}75^{5 x + 3})\)
\(\log(y) = \log(9\,400) + \log(0{,}75^{5 x + 3})\)
1p
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\(\log(y) = \log(9\,400) + (5 x + 3) ⋅ \log(0{,}75)\)
\(\log(y) = \log(9\,400) + 5 x ⋅ \log(0{,}75) + 3 ⋅ \log(0{,}75)\)
1p
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\(\log(y) = 3{,}973... + 5 x ⋅ -0{,}12493... + 3 ⋅ -0{,}12493...\)
\(\log(y) = 3{,}973... - 0{,}62469... ⋅ x - 0{,}37481...\)
Dus \(\log(y) = -0{,}6247 x + 3{,}60\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(\log(y) = 0{,}5705 x + 1{,}69\) in de vorm \(y = b ⋅ g^{x} \text{.}\)
Geef \(b\) in gehelen en \(g\) in twee decimalen.
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\(\log(y) = 0{,}5705 x + 1{,}69\)
\(y = 10^{0{,}5705 x + 1{,}69}\)
1p
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\(y = 10^{0{,}5705 x} ⋅ 10^{1{,}69}\)
\(y = (10^{0{,}5705})^{x} ⋅ 10^{1{,}69}\)
1p
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\(y = 3{,}719...^{x} ⋅ 48{,}977...\)
Dus \(y = 49 ⋅ 3{,}72^{x} \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 3{,}74 ⋅ {}^{4}\!\log(x) - 1{,}25\) in de vorm \(y = {}^{4}\!\log(a x^{b}) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
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\(y = 3{,}74 ⋅ {}^{4}\!\log(x) - 1{,}25\)
\(\text{ } = {}^{4}\!\log(x^{3{,}74}) - 1{,}25\)
1p
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\(\text{ } = {}^{4}\!\log(x^{3{,}74}) + {}^{4}\!\log(4^{-1{,}25})\)
\(\text{ } = {}^{4}\!\log(x^{3{,}74} ⋅ 4^{-1{,}25})\)
1p
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\(\text{ } = {}^{4}\!\log(x^{3{,}74} ⋅ 0{,}176...)\)
Dus \(y = {}^{4}\!\log(0{,}18 ⋅ x^{3{,}74}) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{4}\!\log({69 \over x^{2} \sqrt{x}})\) in de vorm \(y = a + b ⋅ {}^{4}\!\log(x) \text{.}\)
Geef \(a\) in twee decimalen.
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\(y = {}^{4}\!\log({69 \over x^{2} \sqrt{x}})\)
\(\text{ } = {}^{4}\!\log(69 x^{-2{,}5})\)
1p
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\(\text{ } = {}^{4}\!\log(69) + {}^{4}\!\log(x^{-2{,}5})\)
\(\text{ } = {}^{4}\!\log(69) - 2{,}5 ⋅ {}^{4}\!\log(x)\)
1p
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\(\text{ } = 3{,}054... - 2{,}5 ⋅ {}^{4}\!\log(x)\)
Dus \(y = 3{,}05 - 2{,}5 ⋅ {}^{4}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = {}^{2}\!\log(1{,}1 x) - 1{,}7\) in de vorm \(y = a + b ⋅ {}^{3}\!\log(x) \text{.}\)
Geef \(a\) en \(b\) in twee decimalen.
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\(y = {}^{2}\!\log(1{,}1 x) - 1{,}7\)
\(\text{ } = {}^{2}\!\log(1{,}1) + {}^{2}\!\log(x) - 1{,}7\)
1p
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\(\text{ } = {}^{2}\!\log(1{,}1) - 1{,}7 + {{}^{3}\!\log(x) \over {}^{3}\!\log(2)}\)
\(\text{ } = {}^{2}\!\log(1{,}1) - 1{,}7 + {1 \over {}^{3}\!\log(2)} ⋅ {}^{3}\!\log(x)\)
1p
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\(\text{ } = 0{,}137... - 1{,}7 + {1 \over 0{,}630...} ⋅ {}^{3}\!\log(x)\)
\(\text{ } = -1{,}562... + 1{,}584... ⋅ {}^{3}\!\log(x)\)
Dus \(y = -1{,}56 + 1{,}58 ⋅ {}^{3}\!\log(x) \text{.}\)
1p
Herleid tot de gevraagde vorm.
3p
Schrijf de formule \(y = 6 ⋅ \log(4\,000 x) + 5\) in de vorm \(y = a + b ⋅ \log(4 x) \text{.}\)
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\(y = 6 ⋅ \log(4\,000 x) + 5\)
\(\text{ } = 6 ⋅ (\log(1\,000) + \log(4 x)) + 5\)
1p
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\(\text{ } = 6 ⋅ (3 + \log(4 x)) + 5\)
1p
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\(\text{ } = 18 + 6 ⋅ \log(4 x) + 5\)
\(\text{ } = 23 + 6 ⋅ \log(4 x)\)
1p
Druk \(x\) uit in \(y \text{.}\)
3p
\(y = 18 + 2 ⋅ {}^{4}\!\log(6 x - 1)\)
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\(y = 18 + 2 ⋅ {}^{4}\!\log(6 x - 1)\)
\(2 ⋅ {}^{4}\!\log(6 x - 1) = y - 18\)
\({}^{4}\!\log(6 x - 1) = \frac{1}{2} y - 9\)
1p
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\(6 x - 1 = 4^{\frac{1}{2} y - 9}\)
1p
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\(6 x = 4^{\frac{1}{2} y - 9} + 1\)
\(x = \frac{1}{6} ⋅ 4^{\frac{1}{2} y - 9} + \frac{1}{6}\)
1p